A particle is dropped from the top of a tower | vedclass

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Question

(A) Let the total time taken by the particle to reach the ground be $n$ seconds. The initial velocity $u = 0$ and acceleration $a = g = 10 \ m/s^2$.Di

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) Let the total time taken by the particle to reach the ground be $n$ seconds. The initial velocity $u = 0$ and acceleration $a = g = 10 \ m/s^2$.Distance covered in the first $3$ seconds is $S_3 = \frac{1}{2} g(3)^2 = \frac{1}{2} 	imes 10 	imes 9 = 45 \ m$.Distance covered in the last $1$ second is given by $S_{last} = u + \frac{g}{2}(2n - 1) = 0 + \frac{10}{2}(2n - 1) = 5(2n - 1)$.According to the problem,$S_{last} = S_3$,so $5(2n - 1) = 45$.$2n - 1 = 9 \Rightarrow 2n = 10 \Rightarrow n = 5 \ s$.The height of the tower $h = \frac{1}{2} g n^2 = \frac{1}{2} 	imes 10 	imes (5)^2 = 5 	imes 25 = 125 \ m$.

$A$ particle is dropped from the top of a tower. The distance covered by it in the last one second is equal to the distance covered by it in the first three seconds. The height of the tower is $....m$

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