Question

A stone is thrown with an initial speed of $4.9{\mathrm{ms}}^{-1}$ from a bridge in vertically downward direction. It falls down in water after $1$second. Find the height of bridge.

Answer 1

Step1: Given data,

Initial velocity, $\mathrm{u}=4.9{\mathrm{ms}}^{-1}$

Acceleration $=$ acceleration due to gravity (As the stone is falling in the same direction as the direction of acceleration due to gravity, thus, the value of acceleration is positive)

Hence, $\mathrm{a}=\mathrm{g}=9.8{\mathrm{ms}}^{-2}$

Time, $\mathrm{t}=1\mathrm{s}$

Step 2: Calculating the height,

The height of the bridge is calculated by using the second equation of motion,

$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Where, S is the height of the bridge.

Put all the values in equation $\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$, we get

$$\begin{gathered} \mathrm{S}=(4.9m/s)\times (1s)+\frac{1}{2}\times (9.8m/s^2)\times {(1s)}^2 \\ S=4.9+4.9 \\ \mathrm{S}=9.8\mathrm{m} \end{gathered}$$

Hence, the height of the bridge is $9.8\mathrm{m}$.