Question

A particle is projected from ground at angle ${45}^{\circ }$ with initial velocity $20\sqrt{2}m/s.$ Find magnitude of average velocity in a time interval from $t=0$ to $t=3s$. Round off to the nearest integer.

Answer 1

$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$, where $u_x=ucos\theta$ and $u_y=usin\theta$.

$\theta ={45}^o$

$\overrightarrow{u}=20\hat{i}+20\hat{j}\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2\Rightarrow \overrightarrow{s}=(20\hat{i}+20\hat{j})\times 3+\frac{1}{2}(-10\hat{j})\times 9=60\hat{i}+15\hat{j}{\overrightarrow{v}}_{av}=\frac{\overrightarrow{s}}{t}=\frac{60\hat{i}+15\hat{j}}{3}=20\hat{i}+5\hat{j}\left|{\overrightarrow{v}}_{av}\right|=\sqrt{425}=20.6m/s=21m/s$