A target is fixed on the top of a tower `13 m` high. A person standing at a distance of `50 m` from the pole is capable of projecting a stone with a velocity `10 sqrt(g) ms^-1`. If he wants to strike the target in shortest possible time, at what angle should he project the stone ?

Answer

`theta = tan^-1 ((3)/(5))`

Explanation

`13 = 50 tan theta - (g(50)^2)/(2(10 sqrt(g))^2 cos^2 theta)`
`25 tan^2 theta - 100 tan theta + 51 = 0 rArr tan theta = (17)/(5),(3)/(5)`
Time taken will be less for `tan theta = 3//5`.