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

$v=1400\dfrac{\rm m}{\rm s}-gt$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

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

$0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}$

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