Question

A projectile is launched up and to the right over flat, level ground. Its range is 177m, and its maximum elevation above the ground is 354m. What was the angle between its initial velocity and the ground? Ignore aire resistance.

Answer 1

given

range

$X=tv_{0}cos\Theta =177m$

maximum height

$y=\frac{v_{o}^{2}sin^{2}\Theta }{2g}=354m\\ \\ v_{o}^{2}sin^{2}\Theta=354\times 2g\\ \\ v_{o}^{2}sin^{2}\Theta=354\times2 \times 9.8\\ \\ v_{o}sin\Theta=\sqrt{354\times2 \times 9.8}\\\\ v_{o}sin\Theta=83.3m$

the angle

$\frac{v_{o}sin\Theta }{v_{o}cos\Theta }=\frac{83.3}{177}\\ \\ tan\Theta =\frac{83.3}{177}\\ \\ \Theta =tan^{-1}\left ( \frac{83.3}{177} \right )\\ \\ \Theta =25.2^{0}$

in three significant figure'