A baseball is popped straight up into the air and has a hang-time of 6.25s. Determine the height to which the ball rises before it reaches its peak.(Hint : the time to rise to the peak is one half the total hang time )
SohamRoy123:
we need either the mass or the initial and final velocities
Answers
Answered by
74
At the maximum height of the ball, the velocity will be zero. This will also take half the total time, 3.125 s.
Now use the equation...
Δd = vf*t - 1/2at^2
By setting the final time at the moment of the apex of the height, vf becomes 0.
So the equation will become...
Δd = - 1/2at^2
The equation now resembles the change in height from the ground to the maximum height
Δd = - 1/2at^2 = - 1/2 (-9.8 m/s^2) (3.125 s)^2 = 47.8 m
Now use the equation...
Δd = vf*t - 1/2at^2
By setting the final time at the moment of the apex of the height, vf becomes 0.
So the equation will become...
Δd = - 1/2at^2
The equation now resembles the change in height from the ground to the maximum height
Δd = - 1/2at^2 = - 1/2 (-9.8 m/s^2) (3.125 s)^2 = 47.8 m
Thank you.
Similar questions