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Motion in A Plane question

2023 · 8 Apr · Shift 1 · Q59
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  5. /2023 · 8 Apr · Shift 1 · Q59

Motion in A Plane question

2023 · 8 Apr · Shift 1 · Q59

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
Two projectiles A and B are thrown with initial velocities of 40 m/s40 \mathrm{~m} / \mathrm{s}40 m/s and 60 m/s60 \mathrm{~m} / \mathrm{s}60 m/s at angles 30∘30^{\circ}30∘ and 60∘60^{\circ}60∘ with the horizontal respectively. The ratio of their ranges respectively is (g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)(g=10 m/s2)
  1. A
    4:94: 94:9
  2. B
    2:32: \sqrt{3}2:3​
  3. C
    3:2\sqrt{3}: 23​:2
  4. D
    1:11: 11:1
View written solutionFree

Correct answer: A

  1. Use the formula for horizontal range

For a projectile thrown with speed uuu at angle θ\thetaθ, the range is

R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}R=gu2sin2θ​

  1. Range of projectile A

Given for A:

  • uA=40 m/su_A = 40\,\text{m/s}uA​=40m/s
  • θA=30∘\theta_A = 30^\circθA​=30∘

So,

RA=402sin⁡(2×30∘)10R_A = \frac{40^2 \sin(2\times 30^\circ)}{10}RA​=10402sin(2×30∘)​

RA=1600sin⁡60∘10R_A = \frac{1600 \sin 60^\circ}{10}RA​=101600sin60∘​

RA=160⋅32=803R_A = 160\cdot \frac{\sqrt{3}}{2} = 80\sqrt{3}RA​=160⋅23​​=803​

  1. Range of projectile B

Given for B:

  • uB=60 m/su_B = 60\,\text{m/s}uB​=60m/s
  • θB=60∘\theta_B = 60^\circθB​=60∘

So,

RB=602sin⁡(2×60∘)10R_B = \frac{60^2 \sin(2\times 60^\circ)}{10}RB​=10602sin(2×60∘)​

RB=3600sin⁡120∘10R_B = \frac{3600 \sin 120^\circ}{10}RB​=103600sin120∘​

Since,

sin⁡120∘=sin⁡60∘=32\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}sin120∘=sin60∘=23​​

therefore,

RB=360⋅32=1803R_B = 360\cdot \frac{\sqrt{3}}{2} = 180\sqrt{3}RB​=360⋅23​​=1803​

  1. Find the ratio

RA:RB=803:1803R_A : R_B = 80\sqrt{3} : 180\sqrt{3}RA​:RB​=803​:1803​

Cancelling 3\sqrt{3}3​,

RA:RB=80:180=4:9R_A : R_B = 80 : 180 = 4 : 9RA​:RB​=80:180=4:9

  1. Check options
  • A: 4:94:94:9 ✅
  • B: 2:32:\sqrt{3}2:3​
  • C: 3:2\sqrt{3}:23​:2
  • D: 1:11:11:1

So the correct option is A.

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