Question

A baseball is popped straight up into the air and has a hang-time of 6.25 seconds. 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.)

Answer 1

Height is:
\[h=v_0\sin(\theta)t-\frac{ 1 }{ 2 } g t^2\]and peak is reached when\[v_y=0\rightarrow v_0\sin(\theta)-g t=0\rightarrow v_0\sin(\theta)=g t\], then peak height is:\[h_{peak}=g t^2-\frac{ 1 }{ 2 }g t^2=\frac{ 1 }{ 2 }g t^2\] If you do t=3.125 in the previous expression, you get hpeak\[h_{peak}=\frac{ 1 }{ 2 }9.81·3.125^2\]