SSC 10th Science Maharashtra Board: Numerical Problem 1
Chapter 1 (Gravitation) - Question 5 (a)
Note: This is the 1st numerical problem listed in the Gravitation chapter exercise.
An object takes $5\text{ s}$ to reach the ground from a height of $5\text{ m}$ on a planet. What is the value of $g$ on the planet?
Solution:
Given:
Initial velocity ($u$) = $0\text{ m/s}$ (since the object is falling from rest)
Time taken ($t$) = $5\text{ s}$
Displacement/Height ($s$) = $5\text{ m}$To find:
Acceleration due to gravity ($g$) = ?Formula:
According to Newton's second equation of motion:
$$s = ut + \frac{1}{2}at^2$$
Here, acceleration $a$ is the acceleration due to gravity $g$, so:
$$s = ut + \frac{1}{2}gt^2$$Calculation:
Substitute the given values into the formula:$$5 = (0 \times 5) + \frac{1}{2} \times g \times (5)^2$$
$$5 = 0 + \frac{1}{2} \times g \times 25$$
$$5 = \frac{25}{2}g$$
Rearranging the equation to solve for $g$:
$$g = \frac{5 \times 2}{25}$$
$$g = \frac{10}{25}$$
$$g = 0.4\text{ m/s}^2$$
Answer:
The value of acceleration due to gravity ($g$) on the planet is $0.4\text{ m/s}^2$.
