A projectile is projected at an angle of `60^(@)`. If the projectile just clears a cliff `5sqrt3`m high, at a distance of 10m from the point of projection, then calculate the range of the projectile. Also calculate the maximum height and time of flight.

Explanation

`y=14sqrt13`m `x=14m,theta=60^(@)`
By using y`=xtantheta-(x^2tantheta)/R`
i.e `5sqrt3=10tan60^(@)-((100tan60^(@))/R)`
i.e `(100sqrt3)/R=10sqrt3-5sqrt3=5sqrt3`
`:.R=(100)/5=20m`
We know that R`=4Hcottheta`
Hence H`=R/(4cottheta)=(20)/(4cot60)=(5)/(1sqrt3)`
`H=5sqrt3" or "H==8.66m`
Also `R=1/2gT^2cottheta`
i.e `T^2=(2R)/(gcottheta)" "Tsqrt((2xx20xxsqrt3)/(9.8xx1))`
T=2.66s