The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t2) meter and x = 6t meter, where r is in seconds. The velocity of projection is:
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a
8 m/s
b
6 m/s
c
10 m/s
d
not obtained from the data
answer is C.
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Detailed Solution
y = 8t - 5t2vy=dydt=8-10tx = 6tvx=dydt=6Velocity at any time ‘t’ v=vx2+vy2=62+8-10t2Velocity of projection (i.e., velocity at t = 0)= 62+82=10m/s