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A stone is dropped from the top of a tower of height h. After 1s another stone is dropped from a balcony 20 below the top . Both reach the bottom simultaneously. Find value of h . (g=10m/s^2)


3 years ago

## Answers : (1)

11 Points
							Let the time when 1st stone is dropped from a height h be t. So by the formula ,   $T = \sqrt{2h}\div g$ . Let this be equation 1. Now from a height 20m below the top another stone is dropped. So the height from which the 2nd stone has been dropped is (h-20) and given that 1s later. So the time for 2nd stone to fall on the ground is T-1. So by the same formula  $T = \sqrt{2h}\div g$, where in this case Time =( T-1) and Height = (h-20) which is equal to $T -1= \sqrt{2(h-20)}\div g$. This is equation 2. BUt given that both will reach simultaneously so time is same. SO substitute equation 2 in equation 1. After solving these two equations, you will get 8h/g = 25 which implies h = 250/8  that is 31.25m .

3 years ago
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