Anne throws a ball straight up releasing it from her hand with an initial speed of 20 m/s. The ball is at a height of 1.5 m above the ground when it is released. Determine the maximum height of the ball? Calculate the ball’s speed just before it hits the ground?
h=vₒ²/2g=400/2•9.8=20.4 m
H=20.4+1.5=21.9 m
v=sqrt(2gH) = sqrt(2•9.8•21.9) = 20.7 m/s
Thank you so much!
To determine the maximum height of the ball, we need to use the principles of projectile motion. In this case, when the ball reaches its maximum height, its vertical velocity becomes zero.
First, we need to calculate the time it takes for the ball to reach its highest point. We can use the equation of motion:
v_final = v_initial + (acceleration * time)
In this case, the final velocity (v_final) is 0 m/s, the initial velocity (v_initial) is 20 m/s, and the acceleration is the acceleration due to gravity (-9.8 m/s^2).
0 = 20 + (-9.8 * time)
Solving this equation, we get:
time = 20/9.8
Next, we can calculate the maximum height using the following formula:
max_height = (v_initial^2) / (2 * acceleration)
Plugging in the values, we get:
max_height = (20^2) / (2 * 9.8)
Now we can calculate the maximum height:
max_height = 20.41 meters
To calculate the ball's speed just before it hits the ground, we use the equation for vertical velocity:
v_final = v_initial + (acceleration * time)
In this case, the initial velocity (v_initial) is 20 m/s, the acceleration is the acceleration due to gravity (-9.8 m/s^2), and the time is the time it takes for the ball to reach the ground, which is twice the time it took to reach the maximum height.
time = 2 * (20/9.8)
Plugging in the values, we get:
time = 4.08 seconds
Now we can calculate the final velocity:
v_final = 20 + (-9.8 * 4.08)
Solving this equation, we get:
v_final = -20.04 m/s
The negative sign indicates that the velocity is directed in the opposite direction to the initial velocity, which is downward. Hence, the speed just before the ball hits the ground is approximately 20.04 m/s.