Question

A body is projected with a velocity of $60m/s$ at an angle ${30}^0$ to horizontal :

Answer 1

$\text{Initial velocity:}$
$\overrightarrow{v}=v\cos {30}^0\hat{i}+v\sin {30}^0\hat{j}=60\times \frac{\sqrt{3}}{2}\hat{i}+60\times \frac{1}{2}\hat{j}=30\sqrt{3}\hat{i}+30\hat{j}$
$v_y=60\times \sin {30}^0-gt$
$v_x=60\cos {30}^{0}t$
After 2 secs:
$v_y=60\times \frac{1}{2}-10\times 2=20$;$v_x=60\times \frac{\sqrt{3}}{2}=30\sqrt{3}$
Hence, $\overrightarrow{v}=30\sqrt{3}\hat{i}+10\hat{j}$
$s_x=30\sqrt{3}\times 2=60\sqrt{3}$
$s_y=30\times 2-\frac{1}{2}\times 10\times 2^2=40$
$\overrightarrow{s}=s_x\hat{i}+s_y\hat{j}=60\sqrt{3}\hat{i}+40\hat{j}$

After 3 secs:
$v_y=60\times \frac{1}{2}-10\times 3=0$;
$x=60\times \frac{\sqrt{3}}{2}=30\sqrt{3}$
Hence, $\overrightarrow{v}=30\sqrt{3}\hat{i}+0\times \hat{j}=30\sqrt{3}\hat{i}$