How high a building could Superman jump over if he were to leave the ground with a speed of 60.0 m/s at an angle of 75°?

Nov 12, 2015

$171.04 \text{m}$

Explanation:

The vertical component of his velocity is given by:

${V}_{y} = 60 \cos \left(90 - 75\right) = 60 \sin \left(75\right) = 57.95 \text{m/s}$

We can use:

${v}^{2} = {u}^{2} + 2 a s$

$\therefore 0 = {57.9}^{2} + 2 \left(- 9.8\right) s$

$\therefore s = 171.04 \text{m}$