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=(vo/timessinthetha^(2))/(2g)\\</p><p>Range =((vo)^2*(2*\theta ))/(g)\\</p><p>given,\\</p><p>range=8.5* maximum hight\\</p><p>(vo)^2sin(2*\theta )=(8.5*(vo* sin\theta )^2)/(g)\\</p><p>sin(2*\theta )=(8.5*(sin \theta )^2)/(2)\\</p><p>2* sun(\theta )* cos(\theta )=4.25*(sin \theta )^2\\`
`using,</p><p>{sin(2*\theta )=2* sin(\theta)\cos\theta}\\</p><p>2*cos(theta)=4.25* sin(theta)\\</p><p>tan \theta =(2)/(4.25)\\</p><p>tan \theta =0.47\\</p><p>tan\theta=25.2`