Question

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 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 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?

Class - 11th
Chapter - Motion ​

Answer 1

Answer:

$a) \: s = ut - \frac{1}{2} {ut}^{2}$

$s = 0$

$\frac{ {gt}^{2} }{2} = ut$

$t = \frac{2u}{g} = \frac{2 \times 49}{9.8}$

$t=10s$

$b) \: \: Lift \: moving \: with \: Uniform \: speed  = 5m/s</p><p>$

Ball is throughing from inside the lift

Relative velocity of ball = 49m/s

⇒Relative acceleration = 10

Thus, u relative = velocity of bell

$t \: = \frac{ur}{gr}$

$= \frac{2 \times 49}{9.8}$

$t = 10s$

Answer 2

Answer :-

→ Case 1 : Time taken is 10 seconds .

→ Case 2 : Time taken is 10 seconds .

Explanation :-

1st Case :-

• Initial velocity (u) = 49 m/s

• Gravitational acceleration (g) = -9.8 m/s²

• Final velocity (v) = 0

So time taken by the ball to reach the maximum height will be (time of ascent) :-

v = u + gt

⇒ 0 = 49 + (-9.8)t

⇒ -49 = -9.8t

⇒ t = -49/-9.8

⇒ t = 5 sec

Time of ascent = Time of descent

So time taken by the ball to return to the boy's hand will be :-

= Time of ascent + Time of descent

= 5 sec + 5 sec

= 10 sec

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2nd Case :-

In this case, the lift starts moving upwards with a uniform velocity of 5 m/s .

∴ Acceleration of the lift = 0 m/s²

As there is no acceleration, so relative velocity of the ball with respect to the boy will be same i.e. 49 m/s.

Now, if we put the values in the 1st equation of motion, we will get that the time taken by the ball to return to the boy's hand is 10 sec .