Question

A stone is thrown with an initial speed of $4.9\,m{s^{ - 1}}$ from a bridge in the vertically upward direction. It falls down in water after $2\,s$ . The height of the bridge is:
$\begin{gathered}
  \left( A \right)\,\,24.7\,m \\
  \left( B \right)\,\,19.8\,m \\
  \left( C \right)\,\,9.8\,m \\
  \left( D \right)\,\,4.9\,m \\
\end{gathered} $

Answer 1

Hint: From this question, initial speed of the stone and falling time is given. By substituting the values in the displacement formula of the Kinematics equation of motion we can calculate the value of height of the bridge.


Formula Used:
The expression for finding the height of the bridge is
$h = ut + \dfrac{1}{2}g{t^2}$
Where,
$h$ be the height of the bridge, $u$ be the initial velocity of the stone, $t$ be the time, $g$ be the acceleration due to gravity.

Complete step by step answer:
From the question we know that,
initial velocity of the stone $u = - 4.9\,m{s^{ - 1}}$ (stone is thrown upwards so it is negative direction)
time taken by the stone to falls down in water $t = 2\,s$
we know that stone is thrown from a bridge so gravity takes place,
Hence, acceleration due to gravity $g = 9.81\,m{s^{ - 2}}$.
Height of the bridge is given by $h = ut + \dfrac{1}{2}g{t^2}........\left( 1 \right)$
Substituting all the known values in the equation $\left( 1 \right)$
$h = \left( { - 4.9\,m{s^{ - 1}}} \right)\left( {2s} \right) + \dfrac{1}{2}\left( {9.81\,m{s^{ - 2}}} \right){\left( {2s} \right)^2}$
Perform the arithmetic operations in the above equation we get,
$h = \left( { - 9.8\,m} \right) + 19.6\,m$
Subtract the equation, we get
$h = 9.8\,m$.
Hence the height of the bridge is $9.8\,m$.
Thus from the above option, option $\left( C \right)$ is correct.

Note: From this, we note that the stone is thrown upward vertically; it indicates a negative direction. If the stone is thrown downwards it indicates positive direction. So upward direction is taken as negative and downward direction is taken as positive. And it is important to note that we only take displacement into consideration not the total distance.