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(A) Let the total time of motion be $n$ seconds. The initial velocity $u = 0$ and acceleration $g = 10\,m/s^2$.Distance covered in the first $3$ secon

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) Let the total time of motion be $n$ seconds. The initial velocity $u = 0$ and acceleration $g = 10\,m/s^2$.Distance covered in the first $3$ seconds is given by $S_3 = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2} 	imes 10 	imes 3^2 = 45\,m$.Distance covered in the last second of motion 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 total height $h$ is the distance covered in $5$ seconds: $h = \frac{1}{2}gn^2 = \frac{1}{2} 	imes 10 	imes 5^2 = 5 	imes 25 = 125\,m$.

$A$ ball is dropped from the roof of a tower of height $h$. The total distance covered by it in the last second of its motion is equal to the distance covered by it in the first three seconds. The value of $h$ in meters is $(g = 10\,m/s^2)$.

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