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Tema [17]
3 years ago
7

A bullet is fired from a rifle pointed 45 degrees above horizontal. The bullet leaves the muzzle traveling 1400 m/s. How many se

conds does it take the bullet to reach the high point of its trajectory?
Physics
1 answer:
Bogdan [553]3 years ago
5 0

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.

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