At what angle should a body he projected with a velocity `24 ms^(-1)` just to pass over the obstacle `14 m` high at a distance of `24 m` [Take `g = 10ms^(-2)`]

Options

A. `tan theta = 19//5`
B. `tan theta = 1`
C. `tan theta = 3`
D. `tan theta = 2`

Answer

A, B

Explanation

`x = 24 = u costheta.t`
`rArr t = (24)/(24costheta) = (1)/(costheta)`
`y = 14 = u sinthetat - (1)/(2)g t^(2)`
`rArr 14 = (u sintheta)/(costheta) - (5)/(cos^(2)theta) rArr 14 = u tantheta - 5 sec^(2)theta`
`rArr 5tan^(2)theta - 24 tantheta + 19 = 0 rArr tantheta -5 sec^(2)theta`
`rArr 5tan^(2)theta - 24 tantheta + 19 = 0 rArr tantheta = 1, 19//5`. Ans.