Question

A basketball player jumped to rebound a missed shot. Their initial vertical velocity was measured at 3.7m/s. How long was the player in the air and how high did the player jump?

Answer 1

Given that player's initial velocity is 3.7 m/s upward, So

Using 2nd kinematic equation:

d = Vi*t + (1/2)*a*t^2

d = final displacement = 0 m (since he starts from ground and comes back on ground)

Vi = Initial speed of player = 3.7 m/s

t = hang time = ? sec

a = acceleration due to gravity = -9.81 m/sec^2 (-ve sign since g is always downward)

So,

0 = 3.7*t - (1/2)*9.81*t^2

t = 2*3.7/9.81 = 0.754 sec = time for which player was in the air

Now at the max height speed of player will be zero, So

Using 3rd kinematic equation:

Vf^2 = Vi^2 + 2*a*H_max

Vf = final speed of player at max height = 0 m/sec

So,

0 = 3.7^2 - 2*9.81*H_max

H_max = 3.7^2/(2*9.81)

H_max = 0.6977 m = 0.70 m = Max height jumped by player

Let me know if you've any query.