At what angle should a body be projected with a velocity `20 ms^(-1)` just to pass over the obstacle `12 m` high at a horizontal distance of `24 m`? Take `g= 10^-2)`.

Explanation

Refer to Fig. 2 (d) . 24`,
.
Taking horizotal motion of body projected from `O ` from O to A`
we have, `u_(x) =u cos theta =20 cos theta ms^-1)`,
`x_(0) =0, x=24 m, a_(x) =0, t=t (say)`
As `x=x_(0) +u_(x) t + 1/2 a_(x) t_(2)`
:. `24 =0 + ( u cos theta) xx t =(20 cos thera t)`
or `t=(24)/( 20 cos theta) =6/(5 cos theta)`
Taking vertical upward motion of body form (O) to (A)`,
we have : `y_(y) =u sin theta =20 sin theta, y_(0) =0`,
`y=12 m, a_(y) =- 10 ms^(-2), t=t`
As, `y=y_(0)+ u_(y) t +1/2 a_(y) t_(2)`
:. `12 =0 + (20 sin theta) t + 1/2 xx (-10)t^(2)`
or `12 =(20 sin theta (20 sin theta) xx 6/(5 cos theta) 5 xx (6/(5 cos theta))^(2)`
or `12 =24 tan theta =(5 xx 36)/(25 cos^(2) theta`
`=24 tan theta (360/5 sec^(2) theta`
or `12 =24 tan theta -(36)/5 (1+ tan^(2) theta)
or `36 tan ^(2) -120 tan theta + 96 =0`
or `3 tan^(2) theta -10 tan theta + 8 =0`
or ` (tan theta -2) (3 tan theta -4) =0`
or tan theta =2 or 4//3 `
or ` theta 63^(@) 26' or 53^(@) 4'`.