Physics · Forces and Motion
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A boy on a 20 m high cliff drops a stone. One second later, he throws down another stone. Both the stones hit the ground simultaneously. Find the initial velocity of the second stone. (g=10/𝑠 2)
- A
5 m/s2
- B
10 m/s2
- C
15 m/s2
- D
20 m/s2
- E
30 m/s2
In case of drop initial velocity= 0
Time will be given by h=1/2gt2
t=√2h/g =√2×20/10 = 2 sec.
Since both stones hit the ground simultaneously and the 2nd one was dropped 1 sec later, the time taken by the 2nd stone will be t' =2sec -1sec = 1sec .
H'=vi×t+1/2gt'2 =>20= vi×1+½ ×10×1
vi= 15 ms-1.
This velocity is too low for the second stone to cover the distance in the available time.
This value is equivalent to the acceleration due to gravity, not the initial velocity needed.
Correct. This initial velocity allows the second stone to hit the ground at the same time as the first.
This velocity would cause the second stone to hit the ground too early.
This velocity is excessively high, causing the second stone to reach the ground much sooner than the first.
Tagged under Physics · Forces and Motion · 2016