Question

A ball is dropped, at t = 0, from a certain height above the ground, At t=1s another ball is dropped from the same height. the distance between the two balls at t = 3s is (take g= 10 meters per second).

• A. 5m
• B. 20m
• C. 25m
• D. 125m

Answer 1

The correct option is C

25m

Step 1: Given:

1. A ball is dropped, at t = 0, from a certain height above the ground
2. At t = 1s another ball is dropped from the same height

Step 2: From Newton's second equation of motion:

$s=ut+\frac{1}{2}a{t}^2$

Where u is the final velocity, t is time, a is acceleration, and s is the total distance traveled.

Here, u = 0 (ball is dropped), s = height (h) and g is acceleration due to gravity.

$\mathrm{s}=\mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2$

Step 3: Finding the height (h):

Using the above formula we have,

$\mathrm{h}=\frac{1}{2}\times 10\times \left(3^2-2^2\right)$

$\mathrm{h}=5\times \left(9-4\right)=25\mathrm{m}$

Therefore the distance between the two balls at t = 3s is 25 m.

Hence, the correct option is (C).