A ball dropped from the top of a tower falls | vedclass

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(A) Let the total height of the tower be $H$ and the total time taken to reach the ground be $T$. The initial velocity $u = 0$.For the first half of t

Vedclass · Accepted Answer

$14.14$

Vedclass · Answer

(A) Let the total height of the tower be $H$ and the total time taken to reach the ground be $T$. The initial velocity $u = 0$.For the first half of the height $(H/2)$,the time taken is $t_1 = 10 \, s$.Using the equation of motion $s = ut + \frac{1}{2}gt^2$:$\frac{H}{2} = 0 	imes 10 + \frac{1}{2} 	imes g 	imes (10)^2$$\frac{H}{2} = 50g \Rightarrow H = 100g$For the total height $H$,the total time $T$ is given by:$H = 0 	imes T + \frac{1}{2} 	imes g 	imes T^2$$100g = \frac{1}{2} 	imes g 	imes T^2$$T^2 = 200$$T = \sqrt{200} = 10\sqrt{2} \, s$$T = 10 	imes 1.414 = 14.14 \, s$.

$A$ ball dropped from the top of a tower falls the first half of the height of the tower in $10 \, s$. The total time spent by the ball in the air is ......... $s$. [Take $g = 10 \, m/s^2$]

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