Range is same for the angle if projection \(\theta \text { and } 90-\theta\)

\(\text {Time of flight }=\frac{2 u}{g}\)

It is a uniform circular motion. So the direction of velocity changes but not the magnitude. Therefore speed in considered constant. Again magnitude of acceleration does not change.

Use thumb rule

Distance travelled \(s=\frac{1}{2} g t^{2} ; s \alpha t^{2}\). The ratio of distances travelled by a freely falling body will be with ratio 1:4:9:...

For a free falling body, initial vertical downward velocity = 0
For a body thrown horizontally with a velocity, initial vertical downward velocity = 0
Since both having same initial vertical downward velocity they cover equal vertical distance at any instant

Initially velocity has maximum value and at maximum height velocity becomes zero. After that the velocity becomes negative

\(\vec{v}=2 \hat{l}+t^{2} \hat{j}-9 \vec{k}\)
\(\vec{a}=\frac{d \vec{v}}{d t}=2 t \hat{j}\)
\(\text { When } t=0.5 \mathrm{~s}\)
\(a=1 \mathrm{~ms}^{-2}\)

Velocity and acceleration are in the same direction: So speed increases.

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

Mass, length are scalars. Acceleration is a vector but magnitude of acceleration is a scalar

Unit vector specifies only direction
\(\hat{A}=\frac{\vec{A}}{|\vec{A}|} \)
\(\therefore \hat{i}+\hat{j}=\frac{\hat{l}+\hat{j}}{|\hat{l}+\hat{j}|} \)
\(\hat{i} \text { and } \hat{j} \text { are orthogonal components of vectors. }\)
\(\therefore|\hat{i}+\hat{j}|=\sqrt{1^{2}+1^{2}}=\sqrt{1+1}=\sqrt{2}\)

Answers (a), (b) and (c) are all anticlockwiseand answer (d) alone is in the clockwise direction