A body is projected at an angle of `45^(@)` from a point on the ground at a distance of 30 m from the foot of a vertical pole of height 20m. The body just crosses the top of the pole and strikes the ground at a distance s from the foot of the pole on the other side of the pole. Then, s

Options

A. 20 m
B. 30 m
C. 50 m
D. 60 m

Answer

D

Explanation

According to the question, projection of a body shown in the figure below,

Now, from the above figure, let t be the time of crossing of pole,
then, `30 = t u cos 45^(@)` …(i)
and `20 = u sin 45^(@) t - (10)/(2)t^(2)` ...(ii)
Hence, `30sqrt(2) = ut` and
`rArr 20 = (u)/(sqrt(2))t - 5t^(2)`
`rArr t = sqrt(2)s` and `u = 30m//s`
Hence, `R = (u^(2) sin 2 theta)/(g) = (30 xx 30 xx sin 90^(@))/(10) = 90m`
`:.` Distance between pole and the point at which body strike on the ground,
`rArr s = (90 - 30) = 60 m`
So, the correct option is (d).