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TS EAMCET 03 May 2019 Shift 2 Solved Paper

Section: Physics

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Question : 85 of 160

Marks: +1, -0

A ball is projected vertically up from ground. Boy ‘A’ standing at the window of first floor of a nearby building observes that the tune interval between the ball crossing him while going up and the ball crossing him while going down is 2 s. Another boy B standing on the second floor notices that time interval between the ball passing him twice, dining up motion and down motion is 1 s. Calculate the difference between the vertical positions of boy B and boy A. (Assume g = 10 m/

s^2

)

8.45 m
3.75 m
4.25 m
2.50 m

Solution:

The following figure shows the vertical motion of the ball.

From the figure,

v-v_{A} \; \; =-g t

0-v_{A} \; \; =-10(1)

v_{A} \; \; =10 \text{m/s}

since, Boy

B

observed the ball crossing him while up and down motion in

1 s .

So, the time taken by the ball to reach he floor

B

from floor

A

is 0.5 s.

So

s \; \; =v_{A} t-\; \frac{1}{2} g t^{2}

\; \; =10 \times 0.5-\; \frac{1}{2} \times 10 \times (0.5)^{2}

\; \; =5-1.25

\; \; =3.75 \text{m}

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