At what angle with respect to the horizontal, should a projectile be thrown with a velocity of `19.6 m *^(-1)`, to just clear a wall 14.7 m high, at a distance of 19.6m ?

Explanation

Let the angle of projection be `theta` [Fig.2.68]. Hence horizontal component of velocity `=19.6 cos theta m*s^(-1)` and its vertical component `=19.6 sin theta m*s^(-1)`.

Let the time after which the projectile crosses the wall be t . Considering horizontal motion,
`19.6 =19.6 cos theta xxt or, t=sec theta`
For the vertical motion ,
`14.7=19.6 sin theta xxt -1/2 xx9.8 xxt^2`
`or, 14.7 =19.6 sin theta xx sec theta -4.9 sec^2 theta`
`or, 3=4 tan theta -(1+tan^2theta) or, tan^2 theta -4tan theta +4=0`
or, `(tan theta -2)^2 =0 or, tan theta =2 `
`therefore theta =tan ^(-1) 2 =63.4^@` .