The Jump Shot
STEM Topic 30: VOCABULARY
Acceleration: The rate at which an object's velocity changes over time; it could mean speeding up, slowing down, or changing direction.
Components: The individual parts or elements that make up a whole; in this context, referring to the horizontal and vertical parts of an object's initial velocity.
Cosine and Sine: Trigonometric functions that relate the angles of a right triangle to the ratios of its sides.
Descent: The act of moving downward or falling.
Equations: Mathematical expressions stating that two things are equal (e.g., 2 + 3 = 5).
Horizontal: Going from side to side or parallel to the horizon; in a horizontal direction.
Hypotenuse: The side of a right triangle opposite the right angle, considered the longest side.
Inertia: The tendency of an object to resist changes in its state of motion; an object at rest stays at rest, and an object in motion stays in motion unless acted upon by an external force.
Negligible: So small or unimportant that it can be disregarded or considered insignificant.
Projectile: An object that is thrown or launched and moves along a curved path under the influence of gravity.
Sea Level: The average level of the surface of the sea, used as a reference for elevation measurements.
Standardized: Made consistent or uniform according to a specific standard or criterion.
Trigonometry: A branch of mathematics dealing with the relationships between the angles and sides of triangles.
Velocity: The rate of change of an object's position in a particular direction; it includes both speed and direction.
ACCELERATION DUE TO GRAVITY
Acceleration (speeding up) of deceleration (slowing down) refers to a change in velocity (speed).
What is the acceleration due to gravity on Earth's surface?
Answer: The acceleration due to gravity on Earth's surface is approximately 9.81 meters per second squared (m/s2).
What is 9.81 meters in feet? Note: Rounded to the nearest whole foot. 32feet.
What is ”meters per second squared” about?

If an object is dropped from a height of 20 meters, how long will it take to reach the ground?
Use the equation d = 1/2 * g * t2, where d is the distance, g is the acceleration due to gravity (9.81m/s^2)
Answer: Using the equation d = 1/2 * g * t2, where d is the distance , g is the acceleration due to gravity (9.81m/s^2), and t is the time, we can solve for t to get:
20 = 1/2 * 9.81 * t^2 t^2 = 20 / (1/2 * 9.81) t^2 = 4.06
t = 2.02 seconds (rounded to two decimal places)
2. An object is thrown straight up with an initial velocity of 20 meters per second. How high will it go?
Answer: Using the equation vf^2 = vi^2 + 2 * g * d, where vf is the final velocity (0 m/s at the highest point), vi is the initial velocity (20 m/s), g is the acceleration due to gravity (9.81 m/s^2), and d is the distance (the height we want to find), we can solve for d to get:
0 = 20^2 + 2 * (9.81) * d
d = 20.2 meters (rounded to one decimal place)
3. An object is thrown straight up from the top of a building with an initial velocity of 10 meters per second. If the building is 50 meters tall, how long will it take for the object to hit the ground?
Answer: Using the same equation as in problem 3, we can find the time it takes for the object to reach the top of the building:
0 = 10^2 + 2 * (9.81) * d
d = 5.1 meters (rounded to one decimal place)
4. Now, we can use the equation d = vi * t + 1/2 * g * t^2, where vi is the initial velocity (10 m/s), g is the acceleration due to gravity (9.81 m/s^2), d is the distance (50m), and t is the time we want to find. Solving for t, we get:
50 = 10 * t + 1/2 * (9.81) * t^2
t^2  2.04t  5 = 0
Using the quadratic formula, we get:
t = 4.38 seconds (rounded to two decimal places)
5. An object is dropped from a hot air balloon that is rising at a constant velocity of 5 meters per second. How long will it take for the object to hit the ground?
Answer: The acceleration due to gravity is still 9.81 m/s^2, so we can use the same equation as in problem 2 to find the time it takes for the object to fall:
d = 1/2 * g * t^2
Using the same values as in problem 2, but subtracting the initial velocity of the hot air balloon from the distance, we get:
15 = 1/2 * 9.81 * t^2
t^2 = 3.05
t = 1.75 seconds (rounded to two decimal places)
STEM Topic 30: Projectile Motion
Projectile motion is a form of motion in which an object (called a projectile) is launched near the earth's surface, and it moves along a curved path under the action of gravity only. The implication here is that air resistance is negligible, or in any case is being neglected in all of these equations.
The only force of significance that acts on the object is gravity, which acts downward to cause a downward acceleration. Because of the object's inertia, no external horizontal force is needed to maintain the horizontal velocity of the object.
â€‹
In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. This is the principle of compound motion established by Galileo in 1638
â€‹
Initial velocity of parabolic shot
â€‹
The components of initial velocity of a parabolic shot
Initial Velocity
Let the projectile be launched with an initial velocity v0 which can be expressed as the sum of horizontal and vertical components as follows:
v0 = v0x + v0y
If it’s been a few years since you’ve last used trigonometry, sin(e) and cos(ine) can be easily envisioned as sides of a triangle when measured from a specific angle in the example below, the angle located at point A.
The sin (the rise),.is the Opposite side divided by the Hypotenuse.
The cos (the run), is the Adjacent side divided by the Hypotenuse.
If you take a set of stairs, the higher the sin the steeper the steps. The higher the cos the less steep the steps.
The components v0x and v0y be found if the initial launch angle, θ (theta), is known:
v0 x = v0 cos θ
v0y = v0 sin θ
Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is constant, being equal to v0 cos. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to g. The components of the acceleration are:
ax = 0 (this represent the nonimpact of gravity on horizontal motion)
ay = − g (this represents the impact of gravity on vertical motion)
â€‹
Gravity at Sea Level
â€‹
For purposes of weights and measures, a standard gravity value is, denoted g, and g = (32.1740 ft/s2) rounded, for our purposes to 32 ft/s2)
Assuming the standardized value for g and ignoring air resistance, this means that an object falling freely near the Earth's surface increases its velocity by (32 ft/s or about 22 mph) for each second of its descent. Thus, an object starting from rest will attain a velocity of 32 ft/s) after one second, approximately 64 ft/s) after two seconds, and so on.
â€‹
Also, again ignoring air resistance, any and all objects, when dropped from the same height, will hit the ground at the same time.
â€‹
Klay Shoots a Free Throw (Jump Shot)
Projectile motion is a form of motion in which an object (called a projectile), and is frequently a ball, is launched near the earth's surface, and it moves along a curved path under the action of gravity only. The implication here is that air resistance is negligible, or in any case is being neglected in all of these equations. The only force of significance that acts on the object is gravity, which acts downward to cause a downward acceleration. Because of the object's inertia, no external horizontal force is needed to maintain the horizontal velocity of the object.
When NBA player Klay Thompson steps to the free throw line, a number of key variables can influence his shot. Your challenge is to use the 3 key variables and Klay’s stats to figure out the maximum height the ball reaches on its way into the basket to make the shot.
Given Note: US measures (Intl. Std.)
The Acceleration of Gravity: 16t2 (9.8t2)
Initial Vertical Velocity: 24 ft/sec (7.3m/sec)
Klay’s Release Height: 7 feet (2.13m)
(this is the y intercept, it’s when t=0)
h(t) = 16t2 + 24t + 7
AT WHAT TIME(S) DOES THE BALL REACH 10 FEET?

Plan it out. What strategy will you use? Select one or more representations, such as your equation or a graph (found on the last page), to calculate the value(s) of t when the ball reaches a height of 10 feet.
Strategy A:
The height (h) of the ball, in feet, at a given time (t) is represented by the equation:
h(t) = 16t2 + 24t + 7
where 24 is the initial vertical velocity and 7 is the release height
The value for t at 10 feet would occur at two points in time, one on the way up, the other at the hoop.
Substitute 10 feet for h(t): 10 = 16t2 + 24t + 7
Write in standard form: 0 = at2 + bt + c by subtracting 10 from each side:
0 = 16t2 + 24t + (3); where a = 16, b = 24, and c = 3
Solve using the quadratic formula, t = , or complete the square to find two values of t.
2. Solve your problem. Show all your steps. You may use the graph on the last page or show your work in the space below.
â€‹
Strategy A:

Use the quadratic formula:
â€‹

Complete the square: h(t) = 16t2 + 24t + 7
10 = 16t2 + 24t + 7
3 = 16t2 + 24t
3 = 16(t2 + t)
Your solution: (Round your answer to the nearest hundredth.)

The time(s) the ball will reach 10 feet are: 0.14 and 1.36 seconds
AT WHAT TIME DOES THE BALL REACH THE MAXIMUM HEIGHT?
3. Plan it out. What strategy will you use? Select one or more representations, such as your equation or a graph (found on the last page), to calculate the value(s) of t when the ball reaches its maximum height.
Strategy A:
Represented graphically, the equation for height as a function of time, or h(t), is a parabola. Like all parabolas, it is symmetrical, meaning that it has an axis of symmetry that passes through the vertex, or highest point. Since you now know the two values of t when the ball reaches a height of 10 feet, you can find the axis of symmetry by calculating the halfway point, or mean, between these two times. This will give you the value of t when the ball reaches its maximum height.
Strategy B:
You can use the properties of the graph of the equation for h(t) to find the value of t for the vertex or maximum point. For a parabolic function of the form 0 = at2 + bt + c, where a ≠ 0, the value for time (t) is represented by the xcoordinate of the vertex of the parabola.
4. â€‹Solve your problem. Show all your steps. You may use the graph on the last page or show your work in the space below.
â€‹
Your solution: (Round your answer to the nearest hundredth.)

The time the ball will reach the maximum height is: ¾ or .75 seconds
WHAT IS THE MAXIMUM HEIGHT OF THE BASKETBALL?
5. Plan it out. What strategy will you use? Select one or more representations, such as your equation or a graph (found on the last page), to calculate the maximum height the ball will reach on its way to the basket.
Using the value of t, or time, when the ball reaches its maximum height, you can substitute that value into the equation you set up for h(t) to find the height of the ball at that time, or use graphical representation.
6. Solve your problem. Show all your steps. You may use the graph on the last page or show your work in the space below.
â€‹
â€‹
Your solution: (Round your answer to the nearest whole number.)

The maximum height of the basketball will be at: 16 feet
GRAPH YOUR DATA
â€‹
PROBLEM SET 1  Gravity’s Effect on a Dropped Ball
â€‹
â€‹
â€‹
â€‹
â€‹

If a ball is dropped from a height of 10 meters, how long will it take to hit the ground?
Answer: The time it takes for a ball to hit the ground when dropped from a distance is given by the formula t = sqrt(2d/g),
g is the acceleration due to gravity ( 9.8 m/sec2). Height* d = 10
â€‹
*Note: Height is a way of saying distance. It might be indicated by d or h.
sqrt = Square root is the number which, times itself will give you the number that you are seeking.
Find t (Time) t = sqrt(2*10/9.8) = sqrt(2.04sec) = 1.43 seconds
2. If a ball is dropped from a height of 20 meters, what will be its velocity (v), just before hitting the ground? In order to find the velocity, it’s twostep problem. The first step is to determine time (t) in order to use the standard formula for determining velocity due to gravity. v = 9.8 (m/s2) t
t = sqrt(2d/g), = sqrt(2(20)/9.8) = sqrt(4.09) = 2.02
Answer: The velocity of a ball just before hitting the ground when dropped from a height h is given by the formula v = (9.8)2.02, where g is the acceleration due to gravity ( 9.8 m/s2).
Plugging in h = 20, we get v = sqrt(2.02 x 9.8 x 20) = sqrt(392) = 19.80 m/s (rounded to two decimal places).
3. If a ball is dropped from a height of 30 meters, what will be its velocity just before hitting the ground?
Answer: The velocity of a ball just before hitting the ground when dropped from a height h is given by the formula v = sqrt(2gh).
Plugging in h = 30, we get v = sqrt (2 x 30 x 30) = sqrt 588 = 24.25m/sec (rounded to two decimal places).
4. If a ball is dropped from a height of 30 meters, how long does it take to hit the ground?
Answer: Using the formula t = sqrt(2h/g), where h is the height (30m) and g is the acceleration due to gravity (9.8 m/s2), we get t = sqrt(2 x 30/9.8) = 2.47 seconds.
5. What is the velocity of the ball just before it hits the ground, if it was dropped from a height of 30 meters?
Answer: Using the formula v = sqrt(2gh), where h is the height (30m) and g is the acceleration due to gravity (9.8 m/s2), we get v = sqrt(2 x 9.8 x 30) = 34.64 m/s.
6. If a ball is dropped from a height of 40 meters, how long does it take to hit the ground?
Answer: Using the formula t = sqrt(2h/g), where h is the height (40m) and g is the acceleration due to gravity (9.8 m/s2), we get t = sqrt(2*40/9.8) = 3.16 seconds.
7. What is the velocity of the ball just before it hits the ground, if it was dropped from a height of 40 meters?
Answer: Using the formula v = sqrt(2gh), where h is the height (40m) and g is the acceleration due to gravity (9.8 m/s2), we get v = sqrt(2 (9.8/sec2 )40) = 44.29 m/s.
8. If a ball is dropped from a height of 50 meters, how long does it take to hit the ground?
Answer: Using the formula v = sqrt(2gh), where h is the height (50m) and g is the acceleration due to gravity (9.8 m/s2), we get v = sqrt(2 x 9.8 x 50) = 49.01 m/s.
â€‹
PROBLEM SET 2  PROJECTILE MOTION
What is the acceleration due to gravity on Earth's surface?
Answer: The acceleration due to gravity on Earth's surface is approximately 9.81 meters per second squared (m/s^2).
If an object is dropped from a height of 20 meters, how long will it take to reach the ground?
Answer: Using the equation d = 1/2 * g * t^2, where d is the distance (20m), g is the acceleration due to gravity (9.81m/s^2), and t is the time, we can solve for t to get:
20 = 1/2 * 9.81 * t^2
t^2 = 20 / (1/2 * 9.81)
t^2 = 4.06
t = 2.02 seconds (rounded to two decimal places)
An object is thrown straight up with an initial velocity of 20 meters per second. How high will it go?
Answer: Using the equation vf^2 = vi^2 + 2 * g * d, where vf is the final velocity (0 m/s at the highest point), vi is the initial velocity (20 m/s), g is the acceleration due to gravity (9.81 m/s^2), and d is the distance (the height we want to find), we can solve for d to get:
0 = 20^2 + 2 * (9.81) * d
d = 20.2 meters (rounded to one decimal place)
An object is thrown straight up from the top of a building with an initial velocity of 10 meters per second. If the building is 50 meters tall, how long will it take for the object to hit the ground?
Answer: Using the same equation as in problem 3, we can find the time it takes for the object to reach the top of the building:
0 = 10^2 + 2 * (9.81) * d
d = 5.1 meters (rounded to one decimal place)
Now, we can use the equation d = vi * t + 1/2 * g * t^2, where vi is the initial velocity (10 m/s), g is the acceleration due to gravity (9.81 m/s^2), d is the distance (50m), and t is the time we want to find. Solving for t, we get:
50 = 10 * t + 1/2 * (9.81) * t^2
t^2  2.04t  5 = 0
Using the quadratic formula, we get:
t = 4.38 seconds (rounded to two decimal places)
An object is dropped from a hot air balloon that is rising at a constant velocity of 5 meters per second. How long will it take for the object to hit the ground?
â€‹
Answer: The acceleration due to gravity is still 9.81 m/s^2, so we can use the same equation as in problem 2 to find the time it takes for the object to fall:
d = 1/2 * g * t^2
Using the same values as in problem 2, but subtracting the initial velocity of the hot air balloon from the distance, we get:
15 = 1/2 * 9.81 * t^2
t^2 = 3.05
t = 1.75 seconds (rounded to two decimal places)