Question

A particle is fired with velocity $u$ making angle $\theta$ with the horizontal. What is the magnitude of change in velocity when it is at the highest point ?

• A. $ucos\theta$
• B. $u$
• C. $usin\theta$
• D. $(u\cos \theta -u)$

Answer 1

Answer: (C)

$usin\theta$

At the time of projection, the particle's speed is u and the particle's speed at the highest point the particle's speed is $ucos\theta$. So, the change in speed is,

$\Delta u=u-ucos\theta$

Now, the initial velocity is,

$\overrightarrow{u}=ucos\theta \hat{i}+usin\theta \hat{j}$

and the final velocity (at highest point) is,

$\overrightarrow{v}=ucos\theta \hat{i}$

Therefore, the change in velocity is,

$\Delta \overrightarrow{v}=\overrightarrow{u}-\overrightarrow{v}=ucos\theta \hat{i}+usin\theta \hat{j}-ucos\theta \hat{i}=usin\theta \hat{j}$

$\Rightarrow |\Delta \overrightarrow{v}|=usin\theta$