Ncert Exercises & Solutions

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to \(49~\text{m/s}.\) How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of \(5~\text{m/s}\) and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
1. \(5~\text s, 5~\text s\)
2. \(10~\text s, 10~\text s\)
3. \(10~\text s, 5~\text s\)
4. \(5~\text s, 10~\text s\)

Hint: The speed of the ball will be the same when it reaches the boy's hand's but will be moving in a downward direction

Step 1: Required time will be equal to the time of flight of the ball.

Step 2: Find the time of flight using the relation,

T = 2 [\frac{u}{g}] = 2 \times \frac{49}{9.8}

= 10 sec.

In case, the lift starts moving up with a uniform speed of 5 m s^-1, the above calculation remains the same if we take the frame of reference to the lift. The lift is a non-accelerating frame of reference.

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