A ball is thrown with a velocity whose horizontal component is `12 ms^-1` from a point `15 m` above the ground and `6 m` away from a verticlewall `18.75 m` high in such a way so as just to clear the wall. At what time will it reach the ground ? `(g = 10 ms^-2)`.

Answer

3 s

Explanation

`u cos theta = 12,6 = u cos theta t`
`rArr t = 0.5 s` (Fig. S5.61)
`3.75 = u sin theta t - (1)/(2) "gt"^2`
`rArr (15)/(4) = u sin theta xx (1)/(2) - (1)/(2) 10 ((1)/(2))^2 rArr u sin theta = 10`
`-15 = 10 T - (1)/(2) 10 T^2`
`rArr T^2 - 2 T - 3 = 0 rArr T = 3 s`.
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