A ball is thrown from ground level so as to just clear a wall 4m high at a distance of 4 m and falls at a distance of 14 m from the wall. Find the magnitude and direction of the initial velocity.

Options

A. `sqrt(172) m//s`
B. `sqrt(162) m//s`
C. `sqrt(152) m//s`
D. `sqrt(142) m//s`

Explanation

Referring to let P be a point on the trajectory whose co-ordinates are (4, 4). As the ball strikes the ground at a distance 14 metre from the wall, the range is 4 + 14 = 18 metre.
`y=x tan 0 (J x)/(1)`
Here `x=4y, y=4 and R=18`
`4=4 tan 0 (J-4)/(1) =4 tan 0`
`? (1)/(18) (7)/(9)`
or `tan 0 =9//7, sin 0=9// sqrt(130) and cos 0 =7//sqrt(130) u sin 0`
Again `R=(2//g) u^(2) sin 0 cos 0`
`=(2//9.8)xxu^(2) xx (9//sqrt(130))xx(7//sqrt(130))`
`u^(2) =(18xx9.8xxsqrt(30) xx sqrt(130))/(2xx9xx7) =(98xx13)/(7) =182`
`u=sqrt(82) m//s`