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Question

Physics Question on projectile motion

A car starts from rest and accelerates at 5m/s25\, m / s ^{2} At t=4st=4\, s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t=6st =6\, s ? (Take g=10m/s2)\left. g =10\, m / s ^{2}\right)

A

20m/s,5m/s220\, m / s , 5\, m / s ^{2}

B

20m/s,020\, m / s , 0

C

202m/s,020\, \sqrt{2}\, m / s , 0

D

202m/s,10m/s220\, \sqrt{2}\, m / s , 10\, m / s ^{2}

Answer

202m/s,10m/s220\, \sqrt{2}\, m / s , 10\, m / s ^{2}

Explanation

Solution

Velocity of the car at t=4sect =4 \sec is
Vx=at=4×5=20m/sV _{ x }= at =4 \times 5=20 \,m / s
So horizontal velocity =20m/s=20 \,m / s (remain constant)
Vertical velocity at t=6sect =6 \sec
i.e. after 2sec2 sec of free fall
Vy=gt=20m/sV_{y}=g t=20\, m / s
So net velocity =202+202=202m/s=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \,m / s
and once it starts falling acceleration is only ' gg '
i.e., 10m/s210 \,m / s ^{2}