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

Options

A. 4.9 m
B. 9.6 m
C. 19.6 m
D. 14 m

Answer

B

Explanation

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