1.

The equations of motion of a projectile are given by x = 36t metre and 2y = 96t - 9.8r^(2) metre. The angle of projection is :

Answer»

`SIN^(-1)(4/5)`
`sin^(-1)(3/5)`
`sin^(4/3)`
`sin^(-1)(3/4)`

Solution :Here `v_(x) -dx/DT=36, v_(y)=(dy)/(dt)=48-9.8t`
for `t=0v_(y)=48`
`:.v=(36^(2)+(48)^(2))^(1//2)=60`
Now `sintheta=v_(x)/v=48/60=4/5`
`theta=sin^(-1)(4/5)`


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