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Questions & Answers

CBSE

Physics

Grade 11

Gravitation

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

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=10m/s2.

Formula used: sn=u−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 sn=u−g2(2T−1) where T is the total time of flight which is given by 2hg−−−√ and u=0.

Let the total height of tower be ‘h’ and hence, according to the question, sn=36

Putting the values in the equation, we get;

sn=u−g2(2T−1)

0.36h=g2(22hg−−−√−1)

⟹0.072h+1=0.8h−−−−√

Squaring both sides, we get;

(0.072h+1)2=0.8h

⟹0.0722h2+0.144h+1−0.8h=0

⟹0.0722h2−0.656h+1=0

Solving for h, we get;

h=125m

Therefore, the height of the tower is 125 meters.