Class 11 Physics - Chapter Motion in a straight Line NCERT Solutions | A boy standing on a stationary lift (ope

Welcome to the NCERT Solutions for Class 11th Physics - Chapter Motion in a straight Line. This page offers a step-by-step solution to the specific question from Excercise ".$ex_no." , Question 24: a boy standing on a stationary lift open from abo....
Question 24

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?

Answer

Initial velocity of the ball, u = 49 m/s

Acceleration, a = - g = -9.8 m/s2

Case I:

When the lift was stationary, the boy throws the ball. T

aking upward motion of the ball,

Final velocity, v of the ball becomes zero at the highest point.

From first equation of motion, time of ascent (t) is given as:

v =  u + at

t  =  v-u / a

= -49/-9.8 =  5 s

But, the time of ascent is equal to the time of descent.

Hence, the total time taken by the ball to return to the boy's hand = 5 + 5 = 10 s.

 

Case II:

The lift was moving up with a uniform velocity of 5 m/s. In this case, the relative velocity of the ball with respect to the boy remains the same i.e., 49 m/s. Therefore, in this case also, the ball will return back to the boy's hand after 10 s.

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