Physics, asked by pankajbanjare2000, 11 months ago

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)​

Answers

Answered by justinkoushik1011
1

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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