1.

After time 't' the height ‘y ’ of projectile is y = 8t - 5t^(2) and horizontal distance x = 6t if g=10 ms^(-2)1 , velocity of projectile at this instant is :

Answer»

10m/s
8m/s
6m/s
4m/s

Solution :During PROJECTILE’s MOTION
`x=ucosthetat`
`6t=ucosthetatimplies ucostheta=6impliesu_(x)=6ms^(-1)`
Similarly `y=usinthetat-1/2g t^(2)`
`8T - 5t^(2)=usinthetat-1/2g t^(2)`
`implies usintheta=8impliesu_(y)=8MS^(-1)`
Now `u=sqrt(u_(x)^(2)+u_(y)^(2))=10ms^(-1)`


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