A football is kicked straight upward at a speed of 22 m/s. How long will it be in the air?

Question

Hang time formula is used in this

Answer 1

One the the kinematic equations of motion is v_f = v_i + at. I'm unsure if this is the "hang time" equation you mention. It's possible someone has modified this equation to solve for "hang time". But let's start with v_f = v_i + at and go from there. If v_f = v_i + at then we can solve for t (we are being asked for time) by first subtracting v_i from both sides, then dividing by a to get: t = (v_f - v_i)/a

Let's draw a picture:

The acceleration (a) in this case is the acceleration of gravity (also called g) which is 9.8 m/s² but it's pointing downwards so it is -9.8 m/s²

Let's divide the trip of the football in two pieces, going up and coming back down. On the trip up, the final velocity (when the football reaches the top of it's flight) is 0 m/s because it instantaneously stops going up and starts going down.

The trip up will take the same amount of time as the trip down. Additionally, a = g = -9.8 m/s². So to get the "hang time" we can just take the t = (v_f - v_i)/a , multiply by 2, make v_f = 0, and make a = -9.8 m/s². So, if we let "hang tine" be T_H then T_H = 2(-v_i)/-9.8 or T_H = v_i/4.9 (where v_i is given in m/s). A more generic form, not dependent on units, would be T_H = 2(v_i)/g. Perhaps this is the "hang time" formula, I'm not sure, but it works for me.

So T_H = 2v_i/g = 2(22)/9.8 = 4.49 seconds. Considering significant figures, since there are 2 in the "givens", our answer should be rounded to 2 sig figs, so T_H = 4.5 s