Question

The equations of motion of two stones thrown vertically simultaneously are $s_1=19.6t-4.9t^2$and $s_2=9.8t–4.9t^2$ respectively and the maximum height attained by the first one is $h$. When the height of the first stone is maximum, the height of the second stone will be

• A. $\frac{h}{3}$
• B. $2h$
• C. $h$
• D. $0$

Answer 1

The correct option is D

$0$

Step1: Given data

The equation of motion of the first stone is $s_1=19.6t-4.9t^2$

Equation of motion of second stone is $s_2=9.8t–4.9t^2$

Step2: Formula used:

Velocity is the rate of change of displacement.

$v=\frac{ds}{dt}$

To attain maximum height the final velocity will be zero.

Step 3:Determine the time taken by the first stone to reach the maximum height

Thus the time taken by the first stone to attain maximum height

$$\begin{gathered} \frac{d{s}_1}{dt}=\frac{d\left(19.6t-4.9t^2\right)}{dt} \\ {v}_1=19.6-9.8t \\ 0=19.6-9.8t \\ t=2sec. \end{gathered}$$

Step 4: Obtain the height of the second stone

The height attained by the second stone in time $t=2$ will be-

Substituting the value of $t=2$ in equation $s_2=9.8t–4.9t^2$,

$$\begin{gathered} s_2=9.8t–4.9t^2 \\ {s}_2=9·8\times 2-4·9\times 2^2 \\ {s}_2=0m \end{gathered}$$

So, the height of the second stone will be $0m$ when the first stone reached a maximum height.

Hence, option D is the correct answer.