Question

Two balls are dropped to the ground from different heights. One ball is dropped $2s$ after the other but they both strike the ground at the same time. If the first ball takes $5s$ to reach the ground, then the difference in initial heights is (take, $g=10m{s}^{-2}$)

• A. $20m$
• B. $80m$
• C. $170m$
• D. $40m$
• E. $160m$

Answer 1

The correct option is B

$80m$

Step 1: Given Data

Time taken by the first ball $=5s$

Time taken by the second ball $=5-2=3s$

Acceleration due to gravity $g=10m{s}^{-2}$

Let the height of the two balls be $h_1$ and $h_2$ respectively.

Since the balls are dropped from rest, the initial velocity for both the balls $u=0$

Step 2: Formula Used

According to Newton's second equation of motion,

$S=ut+\frac{1}{2}g{t}^2$

Step 3: Calculate the Heights

Therefore, the height of the first ball,

$$\begin{gathered} h_1=0\times 5+\frac{1}{2}\times 10\times 5^2 \\ =5\times 5^2=125m \end{gathered}$$

Similarly, the height of the second ball,

$$\begin{gathered} h_2=0\times 3+\frac{1}{2}\times 10\times 3^2 \\ =5\times 3^2=45m \end{gathered}$$

Step 4: Calculate the height difference

Therefore, the difference between these two heights,

$$\begin{gathered} h_1-h_2=125-45 \\ =80m \end{gathered}$$

Hence, the correct answer is option (B).