Question

A stone moving vertically upwards has its equation of motion $h = 490t – 4.9 {t^2}$. The maximum height reached by the stone is
1) 12250
2) 1225
3) 36750
4) None of these

Answer 1

Hint: The second motion equation is written in symbolic form$s=ut+\dfrac{1}{2}a{{t}^{2}}$, where $s$is initial distance, $u$is initial velocity of an object,$a$ is the acceleration and t is the time taken by the object. Here in this given question height is considered to as initial distance.

Complete answer:
we know that second equation of motion is $s=ut+\dfrac{1}{2}a{{t}^{2}}$ and here in this question we consider maximum height of the stone as$h$.

Given, equation given in this question is$h=490t-4.9{{t}^{2}}$. From this we can derive the value of $u$ and $a$ by comparing second equation of motion and the given equation in the question.
$u=490$ And $a=-9.8$.

Now, putting these value in the third equation of motion that is ${{v}^{2}}-{{u}^{2}}=2as$
For maximum height, $v=o$ thus from the third equation of motion
${{v}^{2}}-{{u}^{2}}=2as$

putting values of which we have derived from the first equation of motion given above
$u=490$ And $a=-9.8$.

And in the place of $s$ in the third equation of motion we will put which is $h$maximum height reached by the stone.
${{0}^{2}}-{{490}^{2}}=2(-9.8)h$
$-{{(490)}^{2}}=-19.6h$
$h={{(490)}^{2}}/19.6$
$h=12,250m$

Therefore the maximum height at which stone can reach will be 12,250 meters. Option (1) is the correct answer for this question.

Note: For tackling above answered question we need to be well versed in the concepts of equation of motion and how these equation derived from it. Deriving a value from one equation and putting it in to other should be done with proper caution to get the desired result.