AIIMS 2015 Physics UNESCO Sites MCQ Question
A ball is dropped from a bridge 122.5 m above a river. After the ball has been falling for 2s, a second ball is thrown straight down after it. What must the initial velocity of the second ball be so that both hit the water at the same time?
40 m/s
55.5 m/s
26.1 m/s
9.6 m/s
Correct Answer
Detailed Explanation
Step 1: Calculate the total time taken by the first ball (t 1
) The first ball is dropped, so its initial velocity (u 1
) is 0 m/s.
s=u 1
t 1
2 1
gt 1 2
122.5=0+ 2 1
(9.8)t 1 2
122.5=4.9t 1 2
t12
= 4.9 122.5
=25⟹t 1
=5 s Step 2: Determine the time available for the second ball (t 2
) The second ball is thrown 2 s later but reaches the water at the exact same moment as the first.
t 2
=t 1
−2 s=5−2=3 s Step 3: Calculate the initial velocity of the second ball (u 2
) The second ball travels the same distance of 122.5 m in 3 s.
s=u 2
t 2
2 1
gt 2 2
122.5=u 2
(3)+ 2 1
(9.8)(3) 2
122.5=3u 2
+4.9(9) 122.5=3u 2
+44.1 3u 2
=122.5−44.1=78.4 u 2
= 3 78.4
≈26.13 m/s This rounds to 26.1 m/s, which matches option (c).
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