Question

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on a level surface. It travels in the air for 13.6 seconds before it strikes the ground 760 m from the shooter. At what angle above the horizontal was the bullet fired? Round to the nearest whole number and include units in your answer Use g= -9.8 m/s2 for the acceleration of gravity.

Answer 1

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

• Δx=760 m
• V₀=87 m/s
• t=13.6 s
• a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$