The height y and the distance x along the horizontal in the plane of the projectile on a certain planet (with no surrounding atmosphere) are given by y = 8t – 5t2 meter and x = 6t meter where t is in seconds. The velocity with which the projectile is projected is-
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a
8 m/s
b
6 m/s
c
10 m/s
d
Data is insufficient
answer is C.
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Detailed Solution
given y=8t-5t2differentiate y with respect to timevy=dydt=8−10tat t=0, vy=8 given x=6tdifferentiate x with respect to timex component of velocity is vx=dxdt=6---(2)v=vx2+vy2substitute eqns (1),(2)v=82+62=10m/s=velocity with which the projectile is projected