a baseball is popped straight up into the air and has a hang time of 6.25 s. determine the height to which the ball rises before it reaches its peak. (hint: the time rise to the peak is one-half the total hang-time.

Question

anonymous · Answer

if total time is 6.25 secconds then time taken to reach peak is 6.25/2=3.125

for total jouney, $$h=0=ut+1/2at ^{2}$$

so $$ut=-1/2at ^{2},        u=-1/2at, u=-1/2(-10)(6.25), u =31.25ms ^{-1}$$

So for half journey upto the peak 

$$h'=ut'+1/2at ^{2}=31.25(3.125)+1/2(-10)(3.125)^{2}=97.66-48.83=48.83m $$

So 48.83 is the peak height

anonymous · Answer

instead of calculating initial velocity in upward direction and then proceeding, start from the top write the equations accordingly. thats quicker and smarter.