Question

A stone is thrown vertically up from a bridge with velocity $3\,m{s^{ - 1}}$ if it strikes the water under the bridge after $2s$ the bridge is at a height of $(g = 10\,m{s^{ - 2}})$
A. $26\,m$
B. $14\,m$
C. $7\,m$
D. $20\,m$

Answer 1

Hint: In order to solve this question, first we will find the time taken by stone to return back to the bridge from where it is thrown upward and then we will use newton equation of motion to calculate the distance covered by stone below the bridge which will be the height of bridge from water surface.

Complete step by step answer:
Let us suppose stone is thrown from point A on bridge upwards with initial velocity as given in question is $u = 3m{s^{ - 1}}$ when it reach its highest point say B its final velocity will be zero $v = 0$ and acceleration due to gravity will be negative as it is moving upwards $g = - 10m{s^{ - 2}}$ and let us assume time taken to reach point B is $t$ . Then using newton’s equation of motion we have,

$v = u + gt$ Here acceleration of stone is acceleration due to gravity.
Putting the values of parameters we get,
$0 = 3 - 10t$
$\Rightarrow t = \dfrac{3}{{10}}s$
Now, since due to symmetry of path followed under free fall of stone from point B to again point A, stone will take same amount of time to return to point A as it take to each point B from point A so, total time consumed by the stone while moving from point A to point B and then returning back to point A under free fall will be
$t' = 2t$
$\Rightarrow v = \dfrac{3}{5}s$

Now as it is given to us that, time taken to hit the water surface which means total time of journey from point A to point C is ${t_{total}} = 2s$ let us assume $t''$ be the time taken by stone from moving point A to point C in return journey under free fall then, this time will be given as
$t'' = {t_{total}} - t'$
$\Rightarrow t'' = 2 - \dfrac{3}{5}$
$\Rightarrow t'' = \dfrac{7}{5}s$

Now, considering journey of stone from point A to point C under free fall, let H be the height between bridge and water surface and initial velocity at point A will be $u = 3m{s^{ - 1}}$ and acceleration will be $g = 10m{s^{ - 2}}$ again using newton equation of motion
$S = ut + \dfrac{1}{2}a{t^2}$ Or
$\Rightarrow H = ut'' + \dfrac{1}{2}gt'{'^2}$
$\Rightarrow H = 3(\dfrac{7}{5}) + \dfrac{1}{2} \times 10 \times {(\dfrac{7}{5})^2}$
$\Rightarrow H = \dfrac{{21}}{5} + \dfrac{{49}}{5}$
$\Rightarrow H = \dfrac{{70}}{5}$
$\therefore H = 14m$

Hence the correct option is B.

Note:It should be remembered that, when a body is thrown up against gravity then, time taken by the body to reach maximum height and time taken to fall from highest point to the point release is always equal and don’t forget to change the sign of acceleration due to gravity while in up direction motion is negative and under free fall the sign of acceleration is positive.