Question

The maximum horizontal range of a ball projected with a velocity of$39.2m/s$ is (take$g=9.8m/s^2$ )

• A. $100m$
• B. $127m$
• C. $157m$
• D. $177m$

Answer 1

The correct option is C

$157m$

Step 1. Given data:

Draw the required diagram for representing the maximum horizontal range of a ball projected with a velocity $u$ at an angle of $\theta$ from the horizontal

Where, the initial velocity of the projected ball$u=39.2m/s$

And acceleration due to gravity, $g=9.8m/s^2$

Step 2. Formula used:

The horizontal range can be expressed as,

$R=\frac{u^2\sin 2\theta }{g}$……………(1)

It is the horizontal distance between the point of the projection to the point where the object touches the ground on the horizontal axis.

Step 3. Calculating the maximum range,

The maximum horizontal range of a ball is projected at an angle of $\theta =45°$ from the horizontal with a velocity of$39.2m/s$.

Substituting the known values in equation (1), we get

$\begin{array}{rcl}R& =& \frac{u^2\sin \left(2\times 45\right)}{g}(\theta =45°,\sin 90=1)\\ & =& \frac{{(39.2)}^2}{9.8}m\\ & =& 156.8m\\ & =& 157m\end{array}$

Thus, the maximum horizontal range of a ball projected with a velocity of$39.2m/s$is$157m$.

Hence, option C is the correct answer.