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Question

From a cliff of 49m high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. Find out the speed with which he threw the second stone.


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Solution

Step 1: Given

Height of the cliff, s=49m

Time taken by the first stone, tseconds

Time taken by the second stone, (t-1)seconds

The initial velocity of stone one, u1=0m/s

Acceleration due to gravity, a=g=9.8m/s

Step 2: Find the velocity

Let v be the final velocity of the first stone,

From the equation of motion, on the first stone,

v2=u12+2asv2=0+2(9.8)(49)v2=960.4v=30.99v31m/s

Step 3: Find the time

Using the equation of motion on the first stone,

v=u+at31=0+9.8tt=3.16s

Therefore, for stone two,

t-1=3.16-1=2.16s

Step 4: Find the initial velocity of the second stone

Using the equation of motion, s=ut+12at2

For stone two,

49=u2(t-1)+12a(t-1)249=u2(2.16)+12(9.8)(2.16)u2=12.1m/s

Hence, he threw the second stone with the speed of 12.1m/s.


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