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Solution - 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 Hand - Kinematic Equations for Uniformly Accelerated Motion

ConceptKinematic Equations for Uniformly Accelerated Motion

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?

Solution 1

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.

Taking 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.

Solution 2

When either the lift is at rest or the lift is moving either vertically upward or downward with a constant speed, we can apply three simple kinetnatic motion equations presuming a = ± g (as the case may be).In present case u = 49 ms-1 (upward) a = g = 9.8 ms-2(downward)

If the ball returns to boy’s hands after a time t, then displacement of ball relative to boy is zero i.e., s = 0. Hence, using equation s = ut + 1/2 at 2, we have

`0 = 49 +- -1/2xx 9.8 xx t^2`

`=>4.9 t^2 - 49t = 0 => t  = 0 or 10 s`

As t = 0 is physically not possible, hence time t = 10s

Is there an error in this question or solution?

APPEARS IN

NCERT Physics Textbook for Class 11 Part 1
Chapter 3: Motion in a Straight Line
Q: 24 | Page no. 59

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Solution for 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 Hand concept: Kinematic Equations for Uniformly Accelerated Motion. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)
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