Question

A projectile is fired in such a way that its horizontal range is equal to 8 times its maximum height. What is the angle of projection? 7.13

Answer 1

solution;

$Projectile maximum height=\frac{vo/timessinthetha^{2}}{2g}\\Range =\frac{(vo)^2\times(2\times\theta )}{g}\\given,\\range=8.5\times maximum hight\\(vo)^2sin(2\times\theta )=\frac{8.5\times(vo\times sin\theta )^2}{g}\\sin(2\times\theta )=\frac{8.5\times(sin \theta )^2}{2}\\2\times sun(\theta )\times cos(\theta )=4.25\times(sin \theta )^2$$using,{sin(2\times\theta )=2\times sin(\theta)\cos\theta}\\2\timescos(theta)=4.25\times sin(theta)\\tan \theta =\frac{2}{4.25}\\tan \theta =0.47\\tan\theta=25.2$