A basketball player jumps straight upward, leaving the ground with an upward initial velocity. In terms of the given information and universal constants, what is the player's vertical jump?

Answer 1

Essentially, what we're supposed to do is "generalize" a 1D kinematics problem. Let's break this down into what we know

v_y0 = velocity the player leaves the ground (we assume we "know" this, but since we don't have a number we'll use the symbol v_y0)

g = acceleration due to gravity = 9.81m/s² (although we know this constant, we will only use the symbol g)

Now, using these constants, we can formulate our velocity and position equations (both are purely in the y-direction)

y(t) = v_y0t - gt²/2

v(t) = v_y0 -gt

These 2 equations are all we'll need to solve this problem.

The question asks for the vertical jump, aka the max height. To find the time at which the max height occurs, set v(t) =0, and solve for t

v(t_maxheight) = 0 = v_y0 -gt_maxheight

⇒ t_maxheight = v_y0/g

And now plug this time into our position

y( t_maxheight) = v_y0 t_maxheight - g t_maxheight² /2 = v_y0( v_y0/g) - g( v_y0/g)²/2 = v_y0²/g - v_y0²/(2g)

y( t_maxheight) = v_y0²/(2g)

Using this equation, we can estimate the max jump height for any player by knowing their vertical jump-speed, v_y0. If you wanted, you could even substitute different values for g to find the jump height on different astral bodies (such as the moon, g_moon ≈ 1.63 m/s² )