Question

A stone is dropped from a building and 2 seconds later another stone is dropped. How far apart are these two stones by the time the first one has reached a speed of 30 ms^-1 (take g= 10 ms^-2)

• A. 80 m
• B. 100 m
• C. 60 m
• D. 40 m

Answer 1

The correct option is D

40 m

Step 1: Given data

Initial velocity U= 0 ms^-1

Final velocity V = 30 ms^-1

Step 2: Calculation

From kinematic equation

$$\begin{gathered} U=v+at \\ U=V-gt \\ 0=30-10\left(t\right) \\ 30=10t \\ t=\frac{30}{10}=3sec \end{gathered}$$

Fro freely falling body distance traveled by first stone,

,$$\begin{gathered} S=Ut+\frac{1}{2}a{t}^2 \\ S=0+\frac{1}{2}g{t}^2 \\ S=\frac{1}{2}g{t}^2 \\ S=\frac{1}{2}\times 10\times 3^2 \\ S=5\times 9 \\ S=45m \end{gathered}$$

For second stone, t= (3 sec- 2 sec) = 1 sec ( because it thrown 2 sec later than the first one)

Distance traveled by second stone,

$$\begin{gathered} S=Ut+\frac{1}{2}a{t}^2 \\ S=0+\frac{1}{2}\times 10\times 1^2 \\ S=5m \end{gathered}$$

The difference in the distance traveled by both the stones,

$45m-5m=40m$

Hence,

Option (d) is correct.