7.
For a particlex, y coordinates, varies with times as
x = 6t, y = 8t - 5t^2. The initial speed of projection is
(x and y are in meter and t in second)
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Answer:
Differentiate w.r.t t
dx/dt=Vx=6
At t=0
Vx= 6m/sec
dy/dt=Vy=8-10t
At t=0
Vy=8m/sec
Intial velocity =(Vx^2+Vy^2)1/2
=(6^2+8^2)^1/2
=10m/sec
Explanation:
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