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

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Question

(B) Let $h$ be the height of the tower and $t$ be the total time taken by the body to reach the ground.Given initial velocity $u = 0$ and acceleration

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let $h$ be the height of the tower and $t$ be the total time taken by the body to reach the ground.Given initial velocity $u = 0$ and acceleration $a = g = 10 \; m/s^2$.The total height is given by $h = \frac{1}{2} g t^2$.The distance covered in the last two seconds is the difference between the total height and the height covered in $(t-2)$ seconds.$40 = \frac{1}{2} g t^2 - \frac{1}{2} g (t-2)^2$.Substituting $g = 10 \; m/s^2$:$40 = 5 t^2 - 5 (t^2 - 4t + 4)$.$40 = 5 t^2 - 5 t^2 + 20 t - 20$.$40 = 20 t - 20$.$60 = 20 t \implies t = 3 \; s$.Now,calculate the height $h$:$h = \frac{1}{2} 	imes 10 	imes (3)^2 = 5 	imes 9 = 45 \; m$.

$A$ body dropped from the top of a tower falls through $40 \; m$ during the last two seconds of its fall. The height of the tower is ..... $m$.

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