What is the initial vertical velocity?

Question

While performing a maximum vertical jump, a 70 kg athlete achieves a maximum vertical jump height of 0.25 m (assume their hand and CoM are displaced vertically an equal amount). What was their initial vertical velocity when they left the floor?

Answer 1

Hi Chris -

To solve this problem, we need to use the 1-dimensional kinematics equation for an object moving with constant acceleration. This is because the athlete is only moving in one direction (the vertical or y-direction), and they are experiencing acceleration at -9.8 m/s² due to the force of gravity. That equation is as follows, first in words and then symbols:

change in vertical position = 1/2 * vertical acceleration * (time)² + initial velocity * time

Δy = ((1/2) * a * t²) + (v_{0 *} t)

The term we are trying to solve for here is v₀, the initial velocity. We choose the moment that the athlete leaves the ground to be our initial moment, and the moment that the athlete reaches their maximum jump height as our final moment - therefore the change in vertical position will be the maximum jump height of 0.25m and the time will be the time from takeoff to maximum height:

0.25 = (1/2) * -9.8 * t² + v₀ * t

We can see that this equation has two unknown in it (v₀ and t), so we need to find another equation involving those variables and then solve both of these as a system of equations.

To find another equation involving v₀ and t, we use another kinematics equation for the change in velocity of an object undergoing acceleration.

v_f = v₀ + at

where a is acceleration and t is time. Since we know the athlete's initial final velocity and acceleration, we can use this equation to solve for t in terms of v₀. For this equation, we will again choose our initial moment to be the moment that the athlete leaves the ground, and our final moment to be the moment they reach their maximum jump height, so their initial velocity will be the same v₀ we had above, and their final velocity will be 0 since they are momentarily at rest at the apex of their jump. This leads to the following, which we solve for t:

0 = v₀ - 9.8t

9.8t = v₀

t = v₀/9.8

Now we can substitute the expression (v₀/9.8) for t in the first equation and solve for v₀:

0.25 = (1/2) * -9.8 * t² + v₀ * t

0.25 = (1/2) * -9.8 * (v₀/9.8)² + v₀ * (v₀/9.8)

0.25 = -4.9(v₀/9.8)² + v₀²/9.8

0.25 = (-1/19.6) v₀² + (1/9.8) v₀²

0.25 = (1/19.6) * v₀²

4.9 = v₀²

v₀ = 2.21 m/s

I hope this helps! Let me know if you have more questions and I'll be happy to explain in more detail.