an object is projected from ground with speed 20m/s at an angle 30 with horizontal .its centripetal acceleration 1s after the projection is

Dear Student,

Please find below the solution to the asked query:

Given that the initial velocity of the particle is 20 m/s and the angle of projection is 30⁰. Let the velocity of the object after t = 1 sec is v. Therefore,

$$\begin{gathered} v \cos \alpha =u \cos \theta \Rightarrow v cos \alpha =20 \cos 30 \\ \Rightarrow v \cos \alpha =10\sqrt{3} \\ \therefore v \sin \alpha =\sqrt{v^2-300} \\ \& \\ v \sin \alpha =u \sin \theta -gt\Rightarrow 20 \sin 30-10\times 1=\sqrt{v^2-300} \\ \Rightarrow v^2-300=0\Rightarrow v=10\sqrt{3} \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

So, the particle will be at its highest point. Therefore the centripetal acceleration is g as shown in the figure below.

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards