A ball is thrown directly upward with an initial velocity of 15 m/s from the rooftop of a building that is 30 meters in height. It is caught by a student whose hand is extended from a window. If the students hand is located 18 meters below the rooftop. determine the balls velocity when it reaches its maximum height in m/s
When the ball reaches its maximum height, its velocity is zero.
The final velocity (Vf) is zero when the ball reaches its maximum height.
The initial velocity (Vi) is 15 m/s.
The acceleration due to gravity (g) is -9.8 m/s^2 (negative because it acts downward).
Using the kinematic equation Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time:
0 = 15 + (-9.8)t
Rearranging the equation to solve for time:
-9.8t = -15
t = -15 / -9.8
t ≈ 1.53 seconds
The ball takes approximately 1.53 seconds to reach its maximum height.
To determine the ball's velocity when it reaches its maximum height, we need to find the velocity at that time.
Using the equation Vf = Vi + at, substituting the values:
Vf = 15 + (-9.8)(1.53)
Vf ≈ 0.094 m/s
Therefore, the ball's velocity when it reaches its maximum height is approximately 0.094 m/s.