A ball projected from ground at an angle of `45^(@)` just clears a wall infront. If point of projection is `4m` from the foot of wall and ball strikes the ground at a distance of `6m` on the other side of the wall, the height of the wall is

Options

A. 4.4 m
B. 2.4 m
C. 3.6 m
D. 1.6 m

Answer

B

Explanation

As ball is projected at an angle `45^@` to the horizontal therefore Range =4H
or 10 =4H `rArr` H=`10/4`=2.5 m (`because` Range =4m + 6m =10m)
Maximum height , `H=(u^2 sin^2 theta)/(2g)`
`therefore u^2=(Hxx2g)/(sin^2 theta) =(2.5xx2xx10)/(1/sqrt2)^2 =100` or , `u=sqrt100=10 ms^(-1)`
Height of wall PA=OA tan `theta -1/2 (g(OA)^2)/(u^2 cos^2 theta)`
`=4-1/2 xx(10xx16)/(10xx10xx1/sqrt2xx1/sqrt2)`=2.4 m