The height y and the distance x along the horizontal plane bf the projectile on a certain planet (with no surrounding atmosphere) are given byy=8t−5t2 metre and x=6t metre where t is in second. The velocity with which the projectile is projected is
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a
8 m/sec
b
6 m/sec
c
10 m/sec
d
Not obtainable from the data given
answer is C.
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Detailed Solution
vy=dydt=8−10t=8 when t = 0 at the time of projectionvx=dxdt=6v=vx2+vy2=(8)2+(6)2=10 m/sec