JEE Main Motion in Plane PYQ

36. (JEE Main 2019 (Online) 12th April Evening Slot )

Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h₁ and h₂. Which of the following is correct ?

A. R² = h₁h₂

B. R² = 16 h₁h₂

C. R² = 4 h₁h₂

D. R² = 2h₁h₂

rrect Option is (B)

The range of two particles are same, that means angle of projections must be complementary to each other.

So one angle = $θ$ and other one is = 90^o - $θ$

R = $\frac{u^{2} \sin 2 θ}{g}$ = $\frac{2 u^{2} \sin θ \cos θ}{g}$

$∴$ R² = $\frac{4 u^{4} \sin^{2} θ \cos^{2} θ}{g^{2}}$

h₁ = $\frac{u^{2} \sin^{2} θ}{2 g}$

h₂ = $\frac{u^{2} \sin^{2} (90^{o} - θ)}{2 g}$ = $\frac{u^{2} \cos^{2} θ}{2 g}$

h₁h₂ = $\frac{u^{4} \sin^{2} θ \cos^{2} θ}{4 g^{2}}$

$\Rightarrow$ h₁h₂ = $\frac{R^{2}}{16}$

$\Rightarrow$ R² = 16 h₁h₂

37. (JEE Main 2019 (Online) 12th April Morning Slot )

A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t₁ and t₂ are the values of the time taken by it to hit the target in two possible ways, the product t₁t₂ is -

A. $\frac{2 R}{g}$

B. $\frac{R}{g}$

C. $\frac{R}{2 g}$

D. $\frac{R}{4 g}$

rrect Option is (A)

Range will be same for time t₁ and t₂, so angles of projection will be ‘ $θ$ ’ & ‘90° – $θ$ ’

$t_{1} = \frac{2 u \sin θ}{g} t_{2} = \frac{2 u \sin (90^{o} - θ)}{g}$

and $R = \frac{u^{2} \sin 2 θ}{g}$

$t_{1} t_{2} = \frac{4 u^{2} \sin θ \cos θ}{g^{2}} = \frac{2}{g} [\frac{2 u^{2} \sin θ \cos θ}{g}]$

= $\frac{2 R}{g}$

38. (JEE Main 2019 (Online) 12th April Morning Slot )

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x² . If it were launched at an angle $θ$ ₀ with speed v₀ then (g = 10 ms^–2) :

A. $θ_{0} = \cos^{- 1} (\frac{1}{\sqrt{5}})$ and $v_{0} = \frac{5}{3}$ ms^-1

B. $θ_{0} = \cos^{- 1} (\frac{2}{\sqrt{5}})$ and $v_{0} = \frac{3}{5}$ ms^-1

C. $θ_{0} = \sin^{- 1} (\frac{2}{\sqrt{5}})$ and $v_{0} = \frac{3}{5}$ ms^-1

D. $θ_{0} = \sin^{- 1} (\frac{1}{\sqrt{5}})$ and $v_{0} = \frac{5}{3}$ ms^-1

rrect Option is (A)

Equation of trajectory is given as
y = 2x – 9x² …(A)

Comparing with equation:
$y = x \tan θ - \frac{g}{2 u^{2} \cos^{2} θ} x^{2}$ ...(B)

We get, $\tan θ = 2$

$∴ \cos θ = \frac{1}{\sqrt{5}}$

Also, $\frac{g}{2 u^{2} \cos^{2} θ} = 9$

$\Rightarrow \frac{10}{2 \times 9 \times {(\frac{1}{\sqrt{5}})}^{2}} = u^{2}; u^{2} = \frac{25}{9}$

$\Rightarrow u = \frac{5}{3} m / s$

39. (JEE Main 2019 (Online) 10th April Evening Slot )

A plane is inclined at an angle $α$ = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms^–1 , from the base of the plane, making an angle $θ$ = 15° with respect to the plane as shown in the figure. the distance from the base, at which the particle hits the plane is close to :
(Take g = 10 ms ^–2) JEE Main 2019 (Online) 10th April Evening Slot Physics - Motion Question 116 English

A. 14 cm

B. 18 cm

C. 20 cm

D. 26 cm

rrect Option is (C)

$T = \frac{2 u \sin θ}{g \cos α}$

$R = u \cos θ T - \frac{1}{2} g \sin α T^{2}$

$= \frac{u \cos θ 2 u \sin θ}{g \cos α} - \frac{g \sin α}{2} \frac{4 u^{2} \sin^{2} θ}{g^{2} \cos^{2} α}$

$= \frac{u^{2} \sin^{2} θ}{g \cos α} - \frac{u^{2} \sin α}{g \cos^{2} α} {1 - \cos 2 θ}$

$= \frac{4 \times \frac{1}{2}}{10 \times \frac{\sqrt{3}}{2}} - \frac{u^{2} \sin α}{g \cos^{2} α} {1 - \frac{\sqrt{3}}{2}}$

$= \frac{4}{10 \sqrt{3}} - \frac{8}{30} {1 - \frac{\sqrt{3}}{2}}$

$= \frac{4}{5 \sqrt{3}} - \frac{8}{30} = \frac{8 \sqrt{3} - 8}{30}$

$= \frac{8 (\sqrt{3} - 1)}{30} = 20 c m$

40. (JEE Main 2019 (Online) 10th January Morning Slot )

Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is -

A. 1 : 16

B. 1 : 8

C. 1 : 2

D. 1 : 4

rrect Option is (A)

JEE Main 2019 (Online) 10th January Morning Slot Physics - Motion Question 126 English Explanation

$R = \frac{u^{2} \sin 2 θ}{g}$

$A = π R^{2}$

$A \propto R^{2}$

$A \propto u^{4}$

$\frac{A_{1}}{A_{2}} = \frac{u_{1}^{4}}{u_{2}^{4}} = {[\frac{1}{2}]}^{4} = \frac{1}{16}$