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 ?

Explanation

From the equtaion of path of projectile,
` y=x tan theta - (gx^2)/(2 u^2 cos^2 theta)`, we have
`13 = 50 tan theta - (g (50)^2)/(2 xx (10 sqrt g)^2 cos^2 theta)`
` = 50 tan theta - (25)/2 sec^2 theta` ,brgt or ` 13 = 50 tan theta - (25)/2 ( 1 + tan^2 theta)`
or ` 25 tan^2 theta- 100 tan theta + 51 =0`
` tan theta = (100 =- sqrt((100)^2 - 4 xx25 xx 51)/(50)`
`= (15)/5 or 3/5`
The horzontal distance coverd in time (t) is given
` by ` x= ucos theta t ` or ` 50 = (10 sqrt g) cos theta t`
` or ` t= (50)/( 10 sqrt g cos theta)`
Hence, (t) will be minimum of the value of ` cso theta` is maximum. It will be so if the value of ` tan theta ` is least
i.e. ` tan theta = 3/5 = 0. 6 ` or ` theta 30^@ 58' `.