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Relative velocity in 1-d

Question

An elevator without a ceiling is ascending up with an acceleration of 5 ms-2 A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of 10 ms-1 and floor of the elevator is at a height of 50 mfrom the ground. The initial speed of the ball is 15 ms-1 with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions. g=10ms-2

Moderate
Question

The time in which the ball strikes the floor of elevator is given by

Solution

Velocity of ball with respect to elevator is 15 m/s (up) and elevator has a velocity of 10 m/s (up). Therefore, absolute velocity of ball is 25 m/s (upwards). Ball strikes the floor of elevator if,

An elevator without a ceiling is ascending up with an acceleration of `5 ms^-2.`  A boy on the elevator shoots a ball in vertical upward direction from -  Sarthaks eConnect | Largest


S1=S2+2   10t+2.5t2=25t-5t2+2
Solving this equation we get,
t=2.13 s

Question

The maximum height reached by ball, as measured from the ground would be

Solution

If the ball does not collide, then it will reach its maximum height in time,
t0=ug=2510=2.5 s
Since, t<t0, therefore as per the question ball is at its maximum height at 2.13 s.
hmax=50+2+25×2.13-5×(2.13)2 =82.56 m

Question

Displacement of ball with respect to ground during its flight would be

Solution

 S=25×2.13-5×(2.13)2 =30.56 m

Question

The maximum separation between the floor of elevator and the ball during its flight would be

Solution

 At maximum separation, their velocities are same 

   25-10t=10+5t     t=1 s
Maximum separation =2+S2-S1
=2+25×1-5×(1)2-10×1+2.5(1)2 =9.5 m  



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From a lift moving upwards with a uniform acceleration a = 2 ms-2, a man throws a ball vertically upwards with a velocity v = 12 ms-1 relative to the lift. The ball comes back to the man after a time t. Find the value of t in seconds. Take g=10ms-2


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