Question

A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many seconds does it take the bullet to reach the high point of its trajectory?

Answer 1

The bullet's vertical velocity at time
`t` is

`v=1400(\rm m)/(\rm s)-gt`

where
`g=9.80(\rm m)/(\mathrm s^2)` is the acceleration due to gravity.

At its highest point, the bullet's vertical velocity is 0, which happens

`0=1400(\rm m)/(\rm s)-gt\implies t=\frac{1400(\rm m)/(\rm s)}g\approx\boxed{142.857\,\mathrm s}`

(or about 140 s, if you're keeping track of significant figures) after being fired.