A machine launches a tennis ball directly upward at 64 feet per second from a platform that is 80 feet high.
what is the maximum height of the tennis ball?
h max = ho + (V^2-Vo^2)/2g
h max = 80 + (0-(64^2))/-64 = 144 Ft
above gnd.
To find the maximum height of the tennis ball, we need to consider the motion of the ball and its vertical velocity, as well as the initial height of the platform.
The motion of the tennis ball can be described by the equation of motion:
h(t) = -16t^2 + vt + h0
Where:
- h(t) represents the height of the ball at time t
- v represents the initial velocity of the ball (64 feet per second, given in the question)
- h0 represents the initial height of the ball (80 feet, given in the question)
- t represents time in seconds
To find the maximum height, we need to determine the time at which the ball reaches its peak. At the maximum height, the vertical velocity of the ball is zero. So we can find the time by setting the derivative of the equation h(t) to zero:
h'(t) = -32t + v = 0
Solving for t:
-32t + 64 = 0
32t = 64
t = 2 seconds
Substituting this value of t back into the equation of motion, we can find the maximum height:
h(2) = -16(2)^2 + 64(2) + 80
h(2) = -64 + 128 + 80
h(2) = 144 feet
Therefore, the maximum height of the tennis ball is 144 feet.