Spud Webb, height 5'7", was one of the shortest basketball players to play in the NBA. But he had an impressive vertical leap; he was reputedly able to jump 110 cm off the ground.
a. To jump this high, with what speed would he leave the ground?
b. For how much time will he be in the air?
c. If he pushes off over a distance of 0.40 m to achieve this vertical leap, what is his acceleration during the pushing-off phase of his jump?

a)   v = 4.64 m / s , b)    t = 0.947 s , c)    t = 0.947 s

Explanation:

We will work on this exercise with vertical launch kinematics, let's start by calculating the height of the jumper in the SI system

         y₀ = 5 ’(0.3048 m / 1’) + 7 ”(2.54 10-2 m / 1”) = 1.70 m

The distance they give is the height of the jump

          y = 1.10 m

Let's use energy conservation

Starting point.  On the floor

          Em₀ = K = ½ m v²

Final point. Maximum height

           Em_{f} = U = m g y

           Em₀ =

           ½ m v² = m g y

           v = √2gy

Let's calculate

          v = √(2 9.8 1.10)

          v = 4.64 m / s

b) Air time is the time to go up plus the time to go down, which is the same

For maximum height the speed is zero

          v = v₀ - g t₁

          t₁ = v₀ / g

          t₁ = 4.64 /9.8

          t₁ = 0.4735 s

The total time is

           t = 2 t₁

           t = 2 0.4735

           t = 0.947 s

c) if it takes a distance of 0.40 to reach speed, what is the acceleration, as it stands on the floor its initial speed is zero

          v² = v₀² + 2 a x

         a = v² / 2x

         a = 4.64²/2 0.40

         a = 26.9 m / s²

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg