A basketball player can jump 1.6 m off the hardwood floor.With what upward velocity did he leave the floor?
A)1.4 m/s
B)2.8 m/s
C)4.2 m/s
D)5.6 m/s

Question

Quizplus · Accepted Answer

We can use the kinematic equation for vertical motion: 
Δy = v0t + 1/2at^2
where Δy is the vertical displacement (1.6 m), v0 is the initial velocity (what we're solving for), a is the acceleration (-9.8 m/s^2), and t is the time in the air. 
We know that at the highest point of the jump, the velocity is 0 m/s. Therefore, we can use the fact that the time to reach the highest point is half the total air time: 
t/2 = sqrt(2Δy/a)
t/2 = sqrt(2(1.6)/(-9.8))
t/2 = 0.357 s
Therefore, the total air time is: 
t = 2(0.357) = 0.714 s
Now we can solve for v0: 
Δy = v0t + 1/2at^2
1.6 = v0(0.714) + 1/2(-9.8)(0.714)^2
v0 = 5.6 m/s

A Basketball Player Can Jump 1