Question

If a football kicked from the ground moves with a velocity of $10m{s}^{–1}$ making an angle of ${60}^o$ with the horizontal, then the difference between its vertical displacement and horizontal displacement after $1s$ would be __________$m.$

• A. $3\sqrt{3-2}$
• B. $5\left(\sqrt{3}-2\right)$
• C. $7\left(\sqrt{3-2}\right)$
• D. $5\left(\sqrt{3-2}\right)$

Answer 1

The correct option is B

$5\left(\sqrt{3}-2\right)$

Step 1:Given data and formula used

Initial velocity $\mathrm{u}=10{\mathrm{ms}}^{-1}$

Angle $\theta ={60}^o$

Time $t=1s$

Here we take a = acceleration due to gravity = $10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$ (

Formula used:

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Step 2:Calculation of the difference between vertical displacement $s_y$ and horizontal displacement $s_x$

On putting the given value in the above formula

$$\begin{gathered} s_x=u_{x}t=5\times 1=5m \\ {s}_y=10\sin \left({60}^0\right)\times 1-\frac{1}{2}10\times {\left(1\right)}^2=10\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times 10\times 1^2=5\sqrt{3}-5=5\left(\sqrt{3}-1\right)m \end{gathered}$$

difference between vertical displacement $s_y$ and horizontal displacement $s_x$

$=s_y-s_x=\left(5\sqrt{3}-5\right)-5=5\left(\sqrt{3}-2\right)m$

Hence, the correct option is (b).