Question

A particle is thrown with an initial speed of $10m{s}^{–1}$ at an angle of $30°$ with the horizontal, under gravity at $t=0s$. At $t=0.5s$, the speed of the particle is

• A. $\begin{array}{l}5\sqrt{3}m/s\end{array}$
• B. $\begin{array}{l}2.5 m/s\end{array}$
• C. $\begin{array}{l}10m/s\end{array}$
• D. $\begin{array}{l}5m/s\end{array}$

Answer 1

The correct option is A

$\begin{array}{l}5\sqrt{3}m/s\end{array}$

Step 1: Given Data

Initial velocity $u=10m{s}^{-1}$

Initial velocity components are $u_x=u\cos \theta$ and $u_y=u\sin \theta$

Angle with the horizontal, $\theta =30°$

Initial time $t=0s$

Final time $t=0.5s$

Let the acceleration be $a$.

Let the final velocity be $v$.

Step 2: Formula Used

The final velocity is given as,

$v=u+at$

Step 3: Velocity along x-axis

Along x-axis the particle does not undergo any acceleration as it travels vertically and only acceleration due to gravity is acting horizontally.

Velocity along the $x$ direction can be given as, at t=0.5 sec

$$\begin{gathered} v_x=u_x+at \\ \Rightarrow v_x=u\cos \theta +at \\ \Rightarrow v_x=10\times \cos 30°+0\times t \\ \Rightarrow v_x=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}m{s}^{-1} \end{gathered}$$

Step 4: Velocity along y-axis

Since the particle is under gravity along the y-axis the particle experiences acceleration due to gravity.

Let the acceleration $a=g=10m{s}^{-2}$

Therefore,

$$\begin{gathered} v_y=u_y+at \\ \Rightarrow v_y=u\sin \theta +at \\ \Rightarrow v_y=10\times \sin 30°-10\times 0.5 \\ \Rightarrow v_y=10\times \frac{1}{2}-5 \\ \Rightarrow v_y=0m{s}^{-1} \end{gathered}$$

Step 5: Resultant Velocity

$$\begin{gathered} v=\sqrt{{v_x}^2+{v_y}^2} \\ =\sqrt{{\left(5\sqrt{3}\right)}^2+0^2} \\ =5\sqrt{3}m{s}^{-1} \end{gathered}$$

Hence, the correct answer is option (A).