Question

From what height did a body fall if it dropped $75{\text{ }}m$ in the last second of its fall? $g = 9.8m/{s^2}$

Answer 1

According to the question, we need to find out the height from which the body fell if it dropped $75{\text{ }}m$ in the last second of its fall. To solve this question, we will use the second equation of motion $s = ut + \dfrac{1}{2}a{t^2}$ .

Complete Step By Step Answer:
In this question, we have to deal with the concept of free fall, so let us first understand what free fall is. Freefall is a common kind of motion which everybody can observe in daily life. If we drop something accidentally we can see its motion. In the beginning, it will have low speed and until the end, it gains speed and before the collision, it reaches its maximum speed. Many factors are there to affect the speed of the object while it is in free fall.
Freefall refers to a situation in physics where the only force acting on an object is gravity and hence acceleration due to gravity. Freefall as its term says is a body falling freely because of the gravitational pull of the earth. This motion will have the effect of acceleration due to gravity. This type of motion will follow the three equations of motion under gravity.
Now, the distance covered by the stone in the nth second is given by the formula,
${s_n} = u + \dfrac{1}{2}a(2n - 1)$
Here u = 0 and a = g,
$s = \dfrac{g}{2} \times (2n - 1)$
By putting the values of s and g in the above equation, we get,
$75 = \dfrac{{9.8}}{2}(2n - 1)$
$75 = 4.9(2n - 1)$ $$
On multiplying 4.9 to the terms in the bracket, we get,
$75 = 9.8n - 4.9$
$9.8n = 79.9$
On taking 9.8 on the other side, we get,
$n = 8.15\sec$
The total time taken in the whole journey=3 seconds and height of the tower is given as,
$s = ut + \dfrac{1}{2}a{t^2}$ $$
Since u = 0 and a = g,
$s = \dfrac{1}{2}g{t^2}$
Now, by putting the values of u and g, we get,
$s = \dfrac{1}{2} \times 9.8 \times 8.15 \times 8.15$
$s = 325.4m$
So, the final answer is $s = 325.4m$ .

Note:
The remarkable observation that all free falling objects fall with the same acceleration was first proposed by Galileo Galilei nearly $400$ years ago. When an object falls under the action of gravity, then it is known as a free fall. In the case of a free fall, the three equations of motion are:
$(1)v = u + gt$
$(2)h = ut + \dfrac{1}{2}g{t^2}$
$(3){v^2} - {u^2} = 2gh$ .