Question #ed9e8

1 Answer
Dec 18, 2014

When the ratio of the range of the projectile to the maximum height is given, the only quantity that can be calculated from this information is the angle of projection. So I rephrased your question.

Projectile Motion :
#v_o# - speed of projection;
#\theta_o# - angle of projection;
#R=(v_o^2\sin2\theta_o)/g# - Range of the projectile,
#H=(v_o^2\sin^2\theta_o)/(2g)# - Maximum height reached by the projectile.

Given that #R=4H# and we are asked to solve for #\theta_o#.

#R=4H \qquad => (v_o^2\sin2\theta_o)/(g) = 4 (v_o^2\sin^2\theta_o)/(2g)#
#\sin2\theta_o=2sin^2\theta_o \qquad => 2\sin\theta_o\cos\theta_o=2\sin^2\theta_o#
#\tan\theta_o=1 \qquad => \theta_o=45^{\circ}#.

So the angle of projection is #45^{\circ}#.