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+1 vote
67.7k views
in Physics by (30.6k points)
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A car starts from rest and accelerates at 5 m/s2. At t = 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 = 6 s ?

(Take g = 10 m/s2)

(1) 20 m/s,5 m/s

(2) 20 m/s, 0 

(3) 20√2 m/s,0 

(4) 20√2 m/s,10m/s2

2 Answers

+2 votes
by (17.0k points)
selected by
 
Best answer

Correct option is (4) 20√2 m/s, 10m/s2

Velocity of the car at t = 4 s,

v = u + at

= 0 + 5 × 4 

= 20 m/s

The ball is dropped at this instant, this means the horizontal velocity of the ball will be vh = 20 m/s.

This velocity will remain constant as there is no horizontal acceleration.

Now, vertical velocity at t = 6 s i.e. Δt = 2 s,

vv = uv + avΔt

= 0 + 10 × 2

= 20 m/s.

Further, the net velocity is,

vnet \(\sqrt{v_h^2 + v_v^2}\)

\(= \sqrt{20^2+20^2}\)

\(= 20\sqrt 2\) m/s

The motion of the ball is under gravity, so,

anet = 10 m/s2

+4 votes
by (30.7k points)

Option : (4). 20√2 m/s,10m/s2

velocity of car at t = 4 sec is 

v = u + at 

v = 0 + 5(4) 

= 20 m/s

At t = 6 sec 

acceleration is due to gravity 

∴ a = g = 10 m/s 

vx = 20 m/s (due to car) 

vy = u + at 

= 0 + g(2) (downward) 

= 20 m/s (downward)

v = \(\sqrt{v^2_x+v^2_y}\)

\(\sqrt{20^2+20^2}\) 

= 20√2 m/s.

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