If `theta` is the angle of projection, R the range, h the maximum height , T the time of flight , then show that (a) tan `theta` = 4h / R and (b) h = `gT^(2)// 8 `

Explanation

(a) Given angle of projection = `theta`
Range , R `= (u^(2) sin 2 theta)/(g) `
`h = h _(max) = (u^(2) sin^(2) theta)/(g)`
Time of flight t `= (2u sin theta)/(g)`
`(h)/(R) = (u^(2) sin^(2) theta)/(2g) xx (g)/(u^(2) sin 2 theta)`
`= (sin^(2)theta)/(2 sin 2 theta) = (sin theta sin theta)/(2 xx 2 sin theta cos theta)`
`:. (h)/(R) = (tan theta)/(4) rArr tan theta = (4h)/(R)`
(b) h ` = (u^(2) sin^(2) theta)/(2g)`
But T `= (2u sin theta)/(g) rArr (u sin theta)/(g) = T // 2 `
`:. h = (u^(2) sin^(2) theta)/(2g) = (u^(2) sin^(2) theta)/(2*g^(2)) * g `
`= ((T)/(2))^(2) xx (1)/(2) = (gT^(2))/(8)`
` :. h = (gT^(2))/(8)` is proved .