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

Options

A. 10m/s
B. 8m/s
C. 6m/s
D. 4m/s

Answer

A

Explanation

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)`