Question

A stone is thrown vertically upwards from the top of a tower $64$ metres high according to the law $S=48t-16t^2$. The greatest height attained by the stone above the ground is

• A. $100m$
• B. $64m$
• C. $36m$
• D. $32m$

Answer 1

The correct option is A

$100m$

Step 1: Given data.

Height of the tower, $h=64m$

Equation of motion of the stone moving vertically upward, $S=48t-16t^2$

Step 2: Finding the greatest height attained by the stone above the ground.

We have,

$S=48t-16t^2$ .…..$\left(i\right)$

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

$\Rightarrow$$\frac{dS}{dt}=\frac{d\left(48t-16t^2\right)}{dt}$

$\Rightarrow$$\frac{dS}{dt}=48-32t$ ……$\left(ii\right)$

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

Since we know that on moving vertically upward the velocity of the stone is zero.

Therefore,

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

$\Rightarrow$$48-32t=0$

$\Rightarrow$ $t=1.5s$

Now,

Substitute the value of $t=1.5s$ in equation $\left(i\right)$ we get,

$\Rightarrow$$S=48\times 1.5-16\times {\left(1.5\right)}^2$

$\Rightarrow$$S=36m$

$\because$ The total height attained by the stone above the ground$=64+36=100m$

Hence, option A is correct.