The equation of motion of a projectile is given by x = 36 t metre and 2y = 96 t – 9.8 t² metre. The angle of projection is:

1. ${\sin }^{-1}\left(\frac{{\displaystyle 4}}{{\displaystyle 5}}\right)$

2. ${\sin }^{-1}\left(\frac{{\displaystyle 3}}{{\displaystyle 5}}\right)$

3. ${\sin }^{-1}\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)$

4. ${\sin }^{-1}\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)$

For a given velocity, a projectile has the same range of R for two angles of projection. If t₁ and t₂ are the times of flight in the two cases then:

(1) $t_1{t}_2\propto R^2$

(2) $t_1{t}_2\propto R$

(3) $t_1{t}_2\propto \frac{1}{R}$

(4) $t_1{t}_2\propto \frac{1}{R^2}$

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

(1) $\sqrt{{\left(v\cos \theta \right)}^2+{\left(v\sin \theta \right)}^2}$

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

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

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

A ball is thrown from a point with a speed v₀ at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v₀/2 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 stone will be:

(1) $\sqrt{\frac{2H}{g}}$

(2) $2\sqrt{\frac{2H}{g}}$

(3) $\frac{2\sqrt{2H\sin \theta }}{g}$

(4) $\frac{\sqrt{2H\sin \theta }}{g}$

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 velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is- 

1. Zero

2. $\frac{1}{\sqrt{2}}m\text{/}{s}^{\text{2}}$ toward north-west

3. $\frac{1}{\sqrt{2}}m\text{/}{s}^{\text{2}}$ toward north-east

4. $\frac{1}{2}m\text{/}{s}^{\text{2}}$ 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: