Question

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°?

Answer 1

`171.04"m"`

Explanation:

The vertical component of his velocity is given by:

`V_y=60cos(90-75)=60sin(75)=57.95"m/s"`

We can use:

`v^2=u^2+2as`

`:.0=57.9^2+2(-9.8)s`

`:.s=171.04"m"`