Question

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the height of tower.​

Answer 1

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$