Question

The stone is thrown vertically upwards and the height $x$ feet reached by the stone in $tsec$, is given by $x=80t-16t^2$. the stone reaches the maximum height in

• A. $2\mathrm{s}$
• B. $2.5\mathrm{s}$
• C. $3\mathrm{s}$
• D. $1.5\mathrm{s}$

Answer 1

The correct option is B

$2.5\mathrm{s}$

Step 1: Given Data:

Equation of motion$\Rightarrow x=80t-16t^2$

The height $x$ feet reached by the stone in $tsec$

Step 2: Formula used:

At maximum height, the velocity will be zero.

We know that velocity is defined as the rate of change of displacement that is

$v=\frac{dx}{dt}$

Where $x$ is displacement, $t$ is time, and $v$ is velocity.

Step 3: Calculation of time at which stone reaches a maximum height:

The equation of motion is given as

$x=80t-16t^2\dots \dots \dots \dots \dots \dots \dots \left(1\right)$

On differentiating the given equation (1) w.r.t. time $t$, we get

$\begin{array}{rcl}x& =& 80t-16t^2\\ \frac{d}{dt}x& =& \frac{d}{dt}\left(80t-16t^2\right)\\ \frac{dx}{dt}& =& 80-32t\dots \dots \dots \dots \dots \dots \dots \left(2\right)\end{array}$

As we know that at maximum height velocity will be zero$\left(i.e.,\frac{dx}{dt}=0\right)$

Substituting $\frac{dx}{dt}=0$ in equation (2)

$\begin{array}{rcl}80-32t& =& 0\\ 80& =& 32t\\ t& =& \frac{80}{32}\\ & =& 2.5s\end{array}$

Hence, option B is the correct answer.