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Question
A stone is dropped from a tower 98 m high. With what speed should a second stone be thrown 1 s later so that both hit the ground at the same time?
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Solution
Height of tower = 98 m
Acceleration due to gravity on stone = 9.8 ms-2.
Initial speed of ball= 0 ms-1.
Let initial speed of second stone is v ms-1.
We know from second equation of motion
S = ut + 1/2 a Xt2.
98 = 0 + 1/2 X9.8Xt2.
t2 = 20
t= 4.47 sec.
As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is = 4.47 - 1 = 3.47sec.
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
98 = vX3.47 + 1/2 X9.8X3.47X3.47.
98 = 3.47Xv + 59
3.47X v = 98 - 59
v= 39/3.47 = 11.23 ms-1.
Initial speed of second stone should be 11.23 ms-1.
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