A winning basketball shot in the last second of a game is launched with an initial speed of 12.2 m/s, 68 degrees above horizontal, and at a height of 2 m above the floor. The ball passes through the basketball hoop mounted at 3.05 m above the floor.
A. What is the ball's velocity at the highest point of travel?
B.What is the maximum height, above the floor, reached by the ball before it arrives at the basket?
C. How much time does the ball take to reach the basket after the ball is thrown?
D. How far from the basket, horizontally, was the ball released when the shot was made?
E. What is the magnitude of the ball's velocity as it passes through the hoop?
Please provide formulas that you use, I am so confused on these types of problems.
To solve this problem, we can break it down into different parts:
A. What is the ball's velocity at the highest point of travel?
To find the velocity at the highest point, we need to consider only the vertical component of the initial velocity. We can use the following formula:
v_vertical = v_initial * sin(theta)
where:
v_vertical is the vertical component of velocity,
v_initial is the initial velocity, and
theta is the launch angle.
In this case, the initial velocity is 12.2 m/s and the launch angle is 68 degrees. Plugging in the values:
v_vertical = 12.2 m/s * sin(68 degrees)
v_vertical ≈ 11.105 m/s
So, the ball's velocity at the highest point of travel is approximately 11.105 m/s vertically.
B. What is the maximum height, above the floor, reached by the ball before it arrives at the basket?
To find the maximum height, we need to consider the vertical motion of the ball. We can use the following formula:
h_max = (v_vertical^2) / (2 * g)
where:
h_max is the maximum height,
v_vertical is the vertical component of velocity, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values:
h_max = (11.105 m/s)^2 / (2 * 9.8 m/s^2)
h_max ≈ 6.141 m
So, the maximum height reached by the ball before it arrives at the basket is approximately 6.141 meters above the floor.
C. How much time does the ball take to reach the basket after the ball is thrown?
To find the time of flight, we need to consider the vertical motion of the ball. We can use the following formula:
t = (2 * v_vertical) / g
where:
t is the time of flight,
v_vertical is the vertical component of velocity, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values:
t = (2 * 11.105 m/s) / 9.8 m/s^2
t ≈ 2.259 s
So, the ball takes approximately 2.259 seconds to reach the basket after it is thrown.
D. How far from the basket, horizontally, was the ball released when the shot was made?
To find the horizontal distance, we need to consider the horizontal motion of the ball. The horizontal velocity remains constant throughout the motion. We can use the following formula:
d_horizontal = v_horizontal * t
where:
d_horizontal is the horizontal distance,
v_horizontal is the horizontal component of velocity, and
t is the time of flight.
To find the horizontal component of velocity, we can use:
v_horizontal = v_initial * cos(theta)
where:
v_horizontal is the horizontal component of velocity,
v_initial is the initial velocity, and
theta is the launch angle.
Plugging in the values:
v_horizontal = 12.2 m/s * cos(68 degrees)
v_horizontal ≈ 4.508 m/s
Now, we can calculate the horizontal distance:
d_horizontal = 4.508 m/s * 2.259 s
d_horizontal ≈ 10.187 m
So, the ball was released approximately 10.187 meters horizontally from the basket.
E. What is the magnitude of the ball's velocity as it passes through the hoop?
To find the magnitude of the velocity as it passes through the hoop, we can use the Pythagorean theorem. The magnitude of velocity is given by:
v = sqrt(v_horizontal^2 + v_vertical^2)
Plugging in the values:
v = sqrt((4.508 m/s)^2 + (11.105 m/s)^2)
v ≈ 11.879 m/s
So, the magnitude of the ball's velocity as it passes through the hoop is approximately 11.879 m/s.