Question

A body falls freely for 10sec. Its average velocity during this journey (take g=$10m{{s}^{-2}}$):
$A.\quad 100m{{s}^{-1}}$
$B.\quad 10m{{s}^{-1}}$
$C.\quad 50m{{s}^{-1}}$
$D.\quad 5m{{s}^{-1}}$

Answer 1

Hint: To solve this problem, we will have to make a few considerations, such as the system has negligible air resistance and the body is stationary initially. The motion equation will be required, given by, $s=ut+\dfrac{1}{2}a{{t}^{2}}$. The average velocity in this case can be found by finding the ratio of the net distance covered to the net time of travel: ${{v}_{avg}}=\dfrac{h}{t}$.

Complete step-by-step solution -
Let’s start by considering a body of mass (m) stationary at a height (h) above the ground. Since the body is stationary, hence the initial velocity of the body is zero. That is, u=0. The acceleration due to gravity (g=$10m{{s}^{-2}}$) is the only accelerating force acting on the body.
Now, we will consider the motion equation, given by: $s=ut+\dfrac{1}{2}a{{t}^{2}}$. Hence, the amount of distance covered by the body in time interval of 10 seconds from the moment, the body starts the free fall is, $h=(0)t+\dfrac{1}{2}(10){{(10)}^{2}}\Rightarrow h=500m$.
Therefore, the body will cover a distance of 500m during the free fall in 10 seconds. The average velocity of the body at the end of 10 seconds can be found out by dividing the total distance covered by the total time taken. That is, ${{v}_{avg}}=\dfrac{h}{t}=\dfrac{500m}{10s}=50m{{s}^{-1}}$.
Hence, the average velocity of the body during the free fall, in the first 10 seconds is ${{v}_{avg}}=50m{{s}^{-1}}$.

Note: Here, we consider an ideal case, that is, the air resistance is negligible and hence, not considered.
Another important result, we find here, in this ideal case is that the average velocity of a body falling freely under the action of gravity in absence of any air resistance will always be equal irrespective of the kind of mass. Hence, the shape of the body or the mass of the body will not affect the final velocity of the body after a time interval t.
If we consider two bodies, let’s take a stone and a paper in vacuum. We need to take the system in vacuum to remove the air resistance, to make it an ideal case. If we consider the same conditions as in the question and try to find the distance travelled by both the stone and the paper at the end of 10 seconds, we will find that, both the stone and the paper will have travelled 500m at the end of 10 seconds.