1.

A projectile is thrown with initial velocity ahat(i) + bhat(j) m/s. If range of projection is twice the maximum height reached by it then :

Answer»

`b=a/2`
b=a
b=2a
b=4a

Solution :Velocity in X-direction `u_(x) = U cos theta= a`
Velocity in Y= direction `u_(y) = u sin theta = b`
Now `R=(u^(2)ssin2theta)/g=(2u^(2)SINTHETACOSTHETA)/g`
`=(2(usintheta)(ucostheta))/g=(2ab)/g`
Also `H_(max)=(usintheta)^(2)/(2g)=b^(2)/(2g)`
`(2ab)/g=(2b^(2))/(2g)`
Thus b=2a


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