Question

A projectile is fired in such a way that its horizontal range is equal to 13.5 times its maximum height. What is the angle of projection?

Answer 1

R tan (theta) =4H

Answer 2

\[R =( u^2 \sin2\theta)/g \] \[H = (u^2 \sin^2\theta)/2g\]

Put R = 13.5 H
You'll get, \[\tan \theta = 4/13.5\]
\[\theta = 16.504\]


Is it correct ?

Answer 3

(R/H) =13.5=4 cotθ
=> θ =16.504

Answer 4

yes it was right!! thank you so much!!!

Answer 5

r=13.5h
=>u^2=20h/sin^2a
=>r=u^2sin a.cos a/10
=>a=cot^(-1)(13.5/4)

Answer 6

:)