1.

The velocity of projection of an oblique projectile is (6hati + 8hatj)ms^(-1) horizontal range of the projectile is:

Answer»

4.9 m
9.6 m
19.6 m
14 m

Solution :Here `v_(x)i=6hati` and `v_(y)hatj=8hatj` and `v_(x)6MS^(-1)`
and `v_(y)=8ms^(-1).`The RESULTANT velocity
`=sqrt((6)^(2)+(8)^(2))=10ms^(-1)`
Now `sintheta=v_(y)/V=8/10=4/5`
and `costheta=v_(x)/v=6/10=3/5`
But `R=(2v^(2)sinthetacostheta)/g`
`=(2xx10^(2))/10xx8/10xx6/10=9.6`m


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