The equation of motion of a projectile is given by \(x = 36 t\) metre and \(2y = 96 t – 9.8 t^2\) metre. The angle of projection is:

1. \(\sin^{− 1} \frac 45\)

2. \(\sin^{− 1} \frac 35\)

3. \(\sin^{− 1} \frac 43\)

4. \(\sin^{− 1} \frac 34\)${}^{}$

For a given velocity, a projectile has the same range of \(R\) for two angles of projection. If \(t_1\) and \(t_2\) are the times of flight in the two cases then:
1. \(t_1t_2 \propto R^2\)

2. \(t_1t_2 \propto R\)

3. \(t_1t_2 \propto \dfrac{1}{R}\)

4. \(t_1t_2 \propto \dfrac{1}{R^2}\)

A body of mass \(m\) is thrown upwards at an angle \(θ\) with the horizontal at velocity \(v\). While rising, the velocity of the mass after \(t\) seconds will be:

1. \(\sqrt{(v\cos\theta)^2+(v\sin\theta)^2}\)

2. \(\sqrt{(v\cos\theta-v\sin\theta)^2-gt}\)

3. \(\sqrt{v^2+g^2t^2-(2v\sin\theta)gt}\)

4. \(\sqrt{v^2+g^2t^2-(2v\cos\theta)gt}\)

A ball is thrown from a point with a speed \(v_0\) at an angle of projection \(θ\). From the same point and at the same instant, a person starts running with a constant speed \(v_0/2\) to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

1. Yes, \(60^°\)

2. Yes, \(30^°\)

3. No

4. Yes, \(45^°\)

A stone is thrown at an angle \(θ\) with the horizontal, reaches a maximum height of \(H\). Then the time of flight of the stone will be:

1. \(\sqrt{\dfrac{2H}{g}}\)

2. \( 2\sqrt{\dfrac{2H}{g}}\)

3.  \(\dfrac{2\sqrt{2H}\sin\theta}{g}\)

4. \(\dfrac{\sqrt{2H}\sin\theta}{g}\)

The maximum horizontal range of a projectile is \(400~\text{m}\). The maximum value of height(ever possible) attained by it will be: 
1. \(100~\text{m}\)
2. \(200~\text{m}\)
3. \(400~\text{m}\)
4. \(800~\text{m}\)

A particle is moving eastwards with a velocity of \(5\ \text{m/s}\). In \(10\) seconds, the velocity changes to \(5\ \text{m/s}\) northwards. The average acceleration in this time is:

1. Zero

2. \(\frac 1{\sqrt2} \ \text{m/s}^2\) toward north-west 

3. \(\frac 1{\sqrt2} \ \text{m/s}^2\) toward north-east

4. \(\frac 1{2} \ \text{m/s}^2\)​​​​​​​ toward north-west

Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

(1) 1, 2, 3, 4

(2) 2, 3, 4, 1

(3) 3, 4, 1, 2

(4) 4, 3, 2, 1

A man standing on a road holds his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically, the speed of raindrops with respect to the road will be:

1. 10 km/hr

2. 20 km/hr

3. 30 km/hr

4. 40 km/hr

Two particles having position vectors \(\overrightarrow{r_{1}} = \left( 3 \hat{i} + 5 \hat{j}\right)\) metres and \(\overrightarrow{r_{2}} = \left(- 5 \hat{i} - 3 \hat{j} \right)\) metres are moving with velocities \(\overrightarrow{v}_{1} = \left( 4 \hat{i} + 3 \hat{j}\right)\)\(\text{m/s}\) and \(\overrightarrow{v}_{2} = \left(\alpha\hat{i} + 7 \hat{j} \right)\)\(\text{m/s}\). If they collide after \(2\) seconds, the value of \(\alpha\) is: