posted by Anonymous on .
A physics student throws a softball straight up into the air. The ball was in the air for a total of 6.16 s before it was caught at its original position.
(a) What was the initial velocity of the ball?
(b) How high did it rise?
Tr + Tf = 6.16 s,
Tf = Tr,
Tr + Tr = 6.16,
2Tr = 6.16,
Tr = 3.08 s. = Rise time or time to reach max. heightt.
a. Vf = Vo + gt,
Vo = Vf - gt,
Vo = 0 - (-9.8)*3.08 = 30.2 m/s.
b. h = Vo*t + 0.5g*t^2,
h = 30.2*3.08 - 4.9*(3.08)^2 = 46.53 m.