A stone moving vertically upwards has its equation of motion $h=490t-4.9t^2$. The maximum height reached by the stone is

The correct option is A

$12250$

Step1: Given data.

Equation of motion of the stone moving vertically upward, $h=490t-4.9t^2$

Step2: Finding the maximum height reached by the stone.

We have,

$h=490t-4.9t^2$ .…..$\left(i\right)$

On differentiating equation $\left(i\right)$ both side we get,

$\Rightarrow$$\frac{dh}{dt}=\frac{d\left(490t-4.9t^2\right)}{dt}$

$\Rightarrow$$\frac{dh}{dt}=490-2\times 4.9t$ ……$\left(ii\right)$

Where $\frac{dh}{dt}$ is the velocity of the stone.

Since, we know that at the maximum height the velocity of the stone is zero.

Therefore,

$\frac{dh}{dt}=0$

$\Rightarrow$$490-2\times 4.9t=0$

$\Rightarrow$ $t=50s$

Now,

Substitute the value of $t=50s$ in equation $\left(i\right)$ we get,

$\Rightarrow$$h=490\times 50-4.9\times {\left(50\right)}^2$

$\Rightarrow$$h=12250m$

Hence, the option A is correct. The maximum height reached by the stone is $12250m$.