A projectile is fired straight upward with an initial velocity of 100 m/s from the top of a building 20 m high and falls to the ground at the base of the building. Find (a) its maximum height above ground; (b) when it passes the top of building; (c) its total time in air.

Question

ybarrap · Answer

Use conservation of energy. Take the potential energy at the top of the building as zero with all energy being kinetic energy. After a while, the ball will reach a maximum height and it's kinetic energy will be zero and its potential energy will be maximum. 

Because energy is conserved, the potential energy at the top must equal the kinetic energy that it started with. 

So, Total energy $ = PE + KE$, where PE $mgh$ and KE $={1 \over 2}mv^2$, where $m$ is mass, $v $ is velocity, $g$ is acceleration due to gravity and $h$ is height.

PE = KE so $\frac 1 2 m v^2=mgh$. So we can solve for $h$ to get the maximum height (hmax)

To get the total time in the air, we can use the formula $ \frac 1 2 a t^2=h$, where h=hmax+20, to account for the fact that we started out 20 m above ground.

I don't know what the second question is asking.