Matt shoots a ball with an initial velocity of 20 m/s at an angle of 53 degrees above the horizontal. Three seconds later, the ball swooshes down a hoop. His friend, Mo, is waiting directly under the basket. Neglect air resistance.
a. What is the speed of the ball when it goes through the hoop?
b. How much higher is the hoop from the release point of the ball?
c. How far is Matt from Mo when he shoots the ball?
d. what is the y-component of the ball's final velocity when it goes through the hoop?
To answer these questions, we can break down the motion of the ball into horizontal and vertical components. Let's go step by step:
a. To find the speed of the ball when it goes through the hoop, we need to find the magnitude of its velocity. The initial velocity has both horizontal and vertical components. The horizontal component (Vx) remains constant throughout, while the vertical component (Vy) changes due to the acceleration due to gravity.
The initial velocity of the ball can be split into horizontal and vertical components using trigonometry. The horizontal component (Vx) is given by Vx = V * cos(theta), where V is the magnitude of the initial velocity (20 m/s) and theta is the angle of 53 degrees.
Vx = 20 m/s * cos(53 degrees)
Vx = 20 m/s * 0.6
Vx = 12 m/s
The vertical component (Vy) is given by Vy = V * sin(theta).
Vy = 20 m/s * sin(53 degrees)
Vy = 20 m/s * 0.8
Vy = 16 m/s
Now, three seconds later, we can find the final vertical velocity (Vy_final) using the equation Vy_final = Vy_initial + (acceleration due to gravity * time).
Since there is no horizontal acceleration and no air resistance, the horizontal velocity remains constant, and the vertical velocity changes by -9.8 m/s (acceleration due to gravity) every second.
Vy_final = 16 m/s + (-9.8 m/s^2 * 3 s)
Vy_final = 16 m/s - 29.4 m/s
Vy_final = -13.4 m/s
The negative sign indicates that the ball is moving downwards. Now we can find the magnitude of the final velocity (V_final) using the Pythagorean theorem.
V_final^2 = Vx^2 + Vy_final^2
V_final^2 = 12 m/s^2 + (-13.4 m/s)^2
V_final^2 = 144 m^2/s^2 + 179.56 m^2/s^2
V_final^2 = 323.56 m^2/s^2
Taking the square root of both sides, we get:
V_final = sqrt(323.56 m^2/s^2)
V_final ≈ 17.99 m/s
Therefore, the speed of the ball when it goes through the hoop is approximately 17.99 m/s.
b. To find how much higher the hoop is from the release point, we need to calculate the vertical displacement (change in height) of the ball.
The vertical displacement (delta_y) can be calculated using the equation delta_y = Vy_initial * t + (1/2) * acceleration due to gravity * t^2.
delta_y = 16 m/s * 3 s + (1/2) * (-9.8 m/s^2) * (3 s)^2
delta_y = 48 m - 44.1 m
delta_y ≈ 3.9 m
Therefore, the hoop is approximately 3.9 meters higher than the release point.
c. To find how far Matt is from Mo when he shoots the ball, we need to calculate the horizontal displacement (range) of the ball.
The horizontal displacement (delta_x) can be calculated using the equation delta_x = Vx_initial * t, where Vx_initial is the horizontal component of the initial velocity and t is the time after the ball is released.
delta_x = 12 m/s * 3 s
delta_x = 36 m
Therefore, Matt is approximately 36 meters away from Mo when he shoots the ball.
d. The y-component of the ball's final velocity when it goes through the hoop (Vy_final) was calculated in part a as -13.4 m/s.