Question

A projectile is launched from ground level with an initial speed of 53.7 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. (Ignore any effects due to air resistance.)

Answer 1

Answer:
`\theta =75.52^(\circ)`

Step-by-step explanation:

Given

initial speed of Launch
`(u)`=53.7 m/s

Range of Projectile =Maximum height of Projectile

Range is given by R
`=(u^2\sin 2\theta )/(g)`

Maximum height is given by
`H_(max)=(u^2\sin ^2\theta )/(2g)`

`R=H_(max)`

`(u^2\sin 2\theta )/(g)=(u^2\sin ^2\theta )/(2g)`

`(u^2\sin 2\theta )/(g)-(u^2\sin ^2\theta )/(2g)=0`

`(u^2\sin \theta )/(g)\cdot \left [ 2\cos \theta -(1)/(2)\right ]=0`

as u cannot be zero therefore

`2\cos \theta -(1)/(2)=0`

`\cos \theta =(1)/(4)`

`\theta =75.52^(\circ)`