If a pro basketball player has a vertical leap of about 25inches, what is his hang time? Use hangtime function v=48t^2

Question

anonymous · Answer

more appropriately this is question for the physics group but: do you mean is hang time the time he stays stationary?

anonymous · Answer

hang time is the time he is in the air

anonymous · Answer

$$\frac{25}{12} = \frac{1}{2}g t^{2}$$
Where g = 32.174. the 25 is over 12 to convert to feet., thus t = :
$$\sqrt{\frac{25(2)}{12g}}$$
Which is .36 secs. This is just the time it takes to reach the 25 in height though. we need to account for the time falling too, which is the same amount of time, so multiply by two to get the final answer of .72 secs.

anonymous · Answer

don't we need to know what his velocity is when he leaps?

anonymous · Answer

At first i thought yes, but if we note that the same time he spends reaching his max height is the same time he spends reaching the floor from that height, then we can just treat this problem as someone dropping something from rest from a height of 25 inches.

anonymous · Answer

ah yes, i see. isn't g unitless though? so don't need to convert to feet?

anonymous · Answer

the g I used is ft per second^2 (as apposed to the 9.81 meters per second^2). Its an acceleration.