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Q.

The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y=8t−5t2 and x=6t . If x and y are measured in metres, the velocity of projection is

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a

6 ms−1

b

14 ms−1

c

8 ms−1

d

10 ms−1

answer is D.

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Detailed Solution

Vy=dydt=8−10t

Vx=dxdt=6

at the time of projection

Vy=dydt=8,Vx=6

V=Vx2+Vy2=62+82=10ms−1

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