AP EAPCET 06-Jul-2022 Shift 2 Paper

The motion of the ball is shown below.
Range, R‌=‌

u2sin‌2θ

g

‌=‌

900sin‌(2×30)

10

=‌

900×sin‌60∘

10

R‌=‌

900×√3

10×2

=‌

900√3

20

=45√3m
Since, the IInd player is at 21‌sqrt‌3‌mathrm‌~m in the direction of kick and he is running to catch the ball, so this can be represented as shown
Let the man runs with speed v towards point P to catch the ball just at point P before it touches the ground in the same time when ball reaches the point P.
The man will travel distance =24√3m
⇒ Distance = speed × time
⇒‌‌2u√3=v×t......(i)
Where, t is the time of run as well as time of flight since both the ball and IInd man start running at same time hence t= time of flight
We know that,
Time of flight, t=‌

2usin‌θ

g

=‌

2×30×sin‌30∘

10

‌t=3 s
Put t=3 s in Eq. (i)
‌2u√3‌=v×3
⇒v‌=8√3m∕ s
Option (c) is the correct answer.