A ball is thrown from the top of a tower with an initial velocity of `10 m//s` at an angle of `30^(@)`above the horizontal. It hits the ground at a distance of 17.3 m from the base of the tower. The height of the tower `(g=10m//s^(2))` will be

Options

A. `(50)/(sqrt(3)) [1-(10)/(sqrt(3))]m`
B. `(50)/(3)[1-(10)/(sqrt(3))]m`
C. `(100)/(sqrt(3)) [1-(10)/(sqrt(3))]m`
D. `(100)/(3)[1-(10)/(sqrt(3))]m`

Answer

A

Explanation

Consider the diagram.

For horizontal motion,
`50 = 10 cos 30^(@) xx t`
`rArr t = (5)/(cos 30^(@)) = (5)/((sqrt(3))/(2)) =(10)/(sqrt(3))`
For vertical motion,
`h = 10 sin 30^(@) xx (10)/(sqrt(3)) - (1)/(2) xx 10 xx ((10)/(sqrt(3)))^(2)`
`= 10 xx (1)/(2) xx (10)/(sqrt(3)) - (100)/(sqrt(3)) (5)/(sqrt(3)) =(100)/(sqrt(3)) ((1)/(2)-(5)/(sqrt(3)))`
`= (50)/(sqrt(3)) [1-(10)/(sqrt(3))] m`