Question

A stone, thrown vertically upward from the surface of the moon at a velocity of $24\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$ reaches a height of $s=24t-0.8t^2$ metre after $t$ second. The acceleration due to gravity in $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$se{c}^2$}\right.$ at the surface of the moon is

• A. $0.8$
• B. $-1.6$
• C. $2.4$
• D. $-4.9$

Answer 1

The correct option is B

$-1.6$

Step1: Given data.

Velocity when thrown vertically upward, $v=24\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Equation of motion of the stone moving vertically upward, $s=24t-0.8t^2$

Step2: Finding the acceleration due to gravity.

We have,

$s=24t-0.8t^2$ .…..$\left(i\right)$

On differentiating equation $\left(i\right)$ both side we get,

$\Rightarrow$$\frac{ds}{dt}=\frac{d\left(24t-0.8t^2\right)}{dt}$

$\Rightarrow$$\frac{ds}{dt}=24-1.6t$ ……$\left(ii\right)$

Where $\frac{ds}{dt}$ is the velocity of the stone.

Again, differentiating equation $\left(ii\right)$ both sides we get,

$\frac{d\left(\frac{dS}{dt}\right)}{dt}=\frac{d\left(24-1.6t\right)}{dt}$

$\Rightarrow$$\frac{d^{2}s}{d{t}^2}=-1.6$

Where $\frac{d^{2}s}{d{t}^2}$ is the acceleration of the stone.

Therefore, the acceleration due to gravity at the surface of the moon is $-1.6\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$se{c}^2$}\right.$.

Hence, option B is correct.