1.

If R is the range and T is the time of flight of a projectile then the angle of projection is given by :

Answer»

`TANTHETA=(G t^(2))/(R)`
`tantheta=(g t^(2))/(2R)`
`tantheta=T^(2)/(Rg)`
`tantheta=T^(2)/(2Rg)`

Solution :`T=(2usintheta)/g` and `R=(2U^(2)sinthetacostheta)/g`
`T^(2)=(4u^(2)sin^(2)THETA)/g:. T^(@)/R=(2sintheta)/(costheta)xx1/g`
or `tantheta=(g t^(2))/(2R)`


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