Question

A particle is thrown with a velocity of $20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ vertically upwards. The distance traveled by the particle in the first second of its descent is calculated by

• A. g
• B. $\frac{g}{2}$
• C. $\frac{g}{4}$
• D. cannot be determined

Answer 1

The correct option is B

$\frac{g}{2}$

Step 1: Given data

A particle is thrown with a velocity, $v=20\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Time for which the body is descended,$t=1s$

Step 2: The characteristics of the particle at the highest point

We know that the velocity of an object at the highest point is equal to zero 0 due to the constant deceleration due to gravity.

At the highest point, the particle comes to the rest position and becomes suspended freely in the air. The particle becomes a freely falling body at the highest point.

Let the velocity at the highest point be the initial velocity, $u$

The acceleration of the particle is equal to the acceleration due to gravity, $g$

Step 3: Finding the distance covered by the particle in the beginning second when it falls freely from the highest point

We know that the distance traveled by a particle is given by,

$$\begin{gathered} s=ut+\frac{1}{2}g{t}^2 \\ Bysubstitutingthecorrespondingvalues,weget \\ s=0\times 1+\frac{1}{2}\times g\times 1^2 \\ s=\frac{g}{2} \end{gathered}$$

Hence, option B is correct.