JEE Main Motion in Plane PYQ

16. (JEE Main 2023 (Online) 24th January Morning Shift )

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is :

A. 136 m

B. 272 m

C. 68 m

D. 192 m

rrect Option is (B)

For vertical throw,

$\begin{aligned} h = \frac{v^{2}}{2 g} \\ v = \sqrt{2 g h} = \sqrt{2 g \times 136} . . . (1) \end{aligned}$

For max range, $θ = 45^{\circ}$

$R_{max} = \frac{v^{2}}{g} . . . (2)$

From (1) and (2)

$\begin{aligned} R_{max} & = \frac{v^{2}}{g} = \frac{2 g \times 136}{g} \\ = 272 m \end{aligned}$

17.(JEE Main 2023 (Online) 11th April Morning Shift )

A projectile fired at $30^{\circ}$ to the ground is observed to be at same height at time $3 s$ and $5 s$ after projection, during its flight. The speed of projection of the projectile is ___________ $m s^{- 1}$ .

(Given $g = 10 {ms}^{- 2}$ )

Correct answer is (80)

Given:

The angle of projection $θ = 30^{\circ}$ .
The projectile is at the same height at time $t_{1} = 3 s$ and $t_{2} = 5 s$ .
The acceleration due to gravity $g = 10 m / s^{2}$ .

We need to find the initial speed of projection, $u$ .

We can use the following equation to find the vertical displacement, $y$ , at any time $t$ :

$y = u_{y} t - \frac{1}{2} g t^{2}$

Where $u_{y}$ is the initial vertical component of the velocity, $u_{y} = u \sin θ$ .

Since the projectile is at the same height at $t_{1}$ and $t_{2}$ , we can write:

$u_{y} t_{1} - \frac{1}{2} g t_{1}^{2} = u_{y} t_{2} - \frac{1}{2} g t_{2}^{2}$

Substitute the values of $t_{1}$ and $t_{2}$ :

$u_{y} (3) - \frac{1}{2} (10) (3)^{2} = u_{y} (5) - \frac{1}{2} (10) (5)^{2}$

Now, let's find the initial vertical component of the velocity, $u_{y}$ :

$u_{y} = u \sin θ = u \sin (30^{\circ}) = \frac{1}{2} u$

Substitute $u_{y}$ in the equation:

$\frac{1}{2} u (3) - \frac{1}{2} (10) (3)^{2} = \frac{1}{2} u (5) - \frac{1}{2} (10) (5)^{2}$

Now, simplify and solve for $u$ :

$3 u - 90 = 5 u - 250$

$2 u = 160$

$u = 80 m / s$

The initial speed of projection is $80 m / s$ .

18.(JEE Main 2023 (Online) 31st January Evening Shift )

Two bodies are projected from ground with same speeds $40 {ms}^{- 1}$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of $60^{\circ}$ , with horizontal then sum of the maximum heights, attained by the two projectiles, is $m$ . (Given $g = 10 {ms}^{- 2}$ )

Correct answer is (80)

Since range is same.

$\begin{aligned} \Rightarrow θ_{1} + θ_{2} = 90^{\circ} \\ \Rightarrow θ_{2} = 30^{\circ} \\ \Rightarrow {(H_{max})}_{1} + {(H_{max})}_{2} = \frac{u^{2} \sin^{2} θ_{1}}{2 g} + \frac{u^{2} \sin^{2} θ_{2}}{2 g} \\ = \frac{40^{2}}{20} (\frac{1}{4} + \frac{3}{4}) = 80 m \end{aligned}$

19. (JEE Main 2022 (Online) 26th July Evening Shift )

Two projectiles are thrown with same initial velocity making an angle of $45^{\circ}$ and $30^{\circ}$ with the horizontal respectively. The ratio of their respective ranges will be :

A. $1 : \sqrt{2}$

B. $\sqrt{2} : 1$

C. $2 : \sqrt{3}$

D. $\sqrt{3} : 2$

rrect Option is (C)

$R = \frac{u^{2} \sin 2 θ}{g}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{\sin 2 θ_{1}}{\sin 2 θ_{2}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

20. (JEE Main 2022 (Online) 26th July Morning Shift )

Two projectiles thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is :

A. $1 : \sqrt{2}$

B. $2 : 1$

C. $\sqrt{2} : 1$

D. $\sqrt{2} : 1$

rrect Option is (C)

$∵$ $t_{a} = \frac{u \sin θ}{g}$

$\Rightarrow \frac{u_{1} \sin (30^{\circ})}{g} = \frac{u_{1} \sin (45^{\circ})}{g}$

$\Rightarrow \frac{u_{1}}{u_{2}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{\sqrt{2}}{1}$