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Question

A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity 20m/s.The second stone will overtake the first after travelling a distance of (g= 10m/s2):

A
13 m
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B
15 m
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C
11.25 m
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D
19.5 m
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Solution

The correct option is C 11.25 m
After time t, distance covered by first stone is s=12gt2
After time (t-1), distance travelled by second stone is s=u(t1)+12g(t1)2=20t20+12g(t1)2
These distance must be equal at the time of overtaking,
Thus, 12gt2=20t20+12g(t1)2
or, 0=20t20+12g(12t)
which gives, 20t20+510t=0 (putting g=10)
or t=1.5
Put this t in equation for distance to get s=12gt2=1210(1.5)2=11.25m

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