Question

A body falls freely from the top of a tower. It covers 36% of the total height in the last second before striking the ground level. The height of the tower is:
A. 50 m
B. 75 m
C. 100 m
D. 125 m

Answer 1

Hint: When a body is dropped from a particular height, it will travel in a straight line from the point through which it is dropped and will fall vertically downwards. This motion is taking place on the surface of earth hence the only force acting on it will be gravitational force or it’s weight. Hence it is an example of accelerated motion where acceleration is $g=10 m/s^2$.
Formula used: $s_n = u-\dfrac g2 (2T-1)$

Complete answer:
Here we have to establish a relation between displacement and height for one second. We know the distance travelled in last second is given by $s_n = u-\dfrac g2 (2T-1)$ where T is the total time of flight which is given by $\sqrt{\dfrac{2h}{g}}$ and u=0.
Let the total height of tower be ‘h’ and hence, according to the question, $s_n = 36%h = 0.36h$
Putting the values in the equation, we get;
$s_n = u-\dfrac g2 (2T-1)$
$0.36h = \dfrac g2 (2\sqrt{\dfrac{2h}{g}}-1)$
$\implies 0.072 h + 1 = \sqrt{0.8 h}$
Squaring both sides, we get;
$(0.072 h +1)^2 = 0.8h$
$\implies 0.072^2h^2 +0.144 h + 1 - 0.8h =0$
$\implies 0.072^2 h^2 - 0.656h +1=0$
Solving for h, we get;
$h=125 m$
Therefore, the height of the tower is 125 meters.

Note:
Student must not get confused by negative and positive signs. We take starting points as origin and then measure quantities with respect to this point. If the direction of measurement of a quantity is towards down or left, then we take negative signs otherwise positive. For example, if the answer would be -125 m, it means that the displacement is downwards with respect to the place where we are standing. In other words, the displacement is in the direction of acceleration due to gravity. In this question, we’ve used direct relation for the displacement in the last second. But one can also proceed by calculating the distance for ‘n’ seconds and then for ‘n-1’ seconds and then subtract them, the results will be the same.