a stone is dropped from the top of a tall cliff and 1 s later a second stone is thrown vertically downward with a velocity of 20 m/s. how far below the top of the Cliff will the second stone overtake the first?
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Dear friend,
The solution is provided below :-
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Answered by
32
Hey dear,
◆ Answer-
s = 11.25 m
◆ Explanation-
Let s be distance in which 2nd stone catches up with first stone and t be time taken.
For 1st stone-
s = ut + 1/2 g(t+1)^2
s = 0×t + 1/2 × 10(t^2+2t+1)
s = 5(t2+2t+1) ...(1)
For 2nd stone-
s = u't + 1/2 gt^2
s = 20t + 1/2 × 10t^2
s = 20t + 5t^2 ...(2)
Equalizing both (1) & (2),
5(t2+2t+1) = 20t + 5t^2
10t + 5 = 20t
t = 10/5
t = 0.5 s
To calculate the distance -
s = 20t + 5t^2
s = 20×0.5 + 5×(0.5)^2
s = 10 + 1.25
s = 11.25 m
Therefore, 2nd stone catches 1st one at 11.25 m from above.
Hope this is helpful...
◆ Answer-
s = 11.25 m
◆ Explanation-
Let s be distance in which 2nd stone catches up with first stone and t be time taken.
For 1st stone-
s = ut + 1/2 g(t+1)^2
s = 0×t + 1/2 × 10(t^2+2t+1)
s = 5(t2+2t+1) ...(1)
For 2nd stone-
s = u't + 1/2 gt^2
s = 20t + 1/2 × 10t^2
s = 20t + 5t^2 ...(2)
Equalizing both (1) & (2),
5(t2+2t+1) = 20t + 5t^2
10t + 5 = 20t
t = 10/5
t = 0.5 s
To calculate the distance -
s = 20t + 5t^2
s = 20×0.5 + 5×(0.5)^2
s = 10 + 1.25
s = 11.25 m
Therefore, 2nd stone catches 1st one at 11.25 m from above.
Hope this is helpful...
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