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Motion in A Plane question

2023 · 10 Apr · Shift 2 · Q45
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  5. /2023 · 10 Apr · Shift 2 · Q45

Motion in A Plane question

2023 · 10 Apr · Shift 2 · Q45

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
Two projectiles are projected at 30∘30^{\circ}30∘ and 60∘60^{\circ}60∘ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:
  1. A
    1:31: \sqrt{3}1:3​
  2. B
    3:1\sqrt{3}: 13​:1
  3. C
    1 : 3
  4. D
    2:32: \sqrt{3}2:3​
View written solutionFree

Correct answer: C

  1. Formula for maximum height of a projectile

For a projectile launched with speed uuu at angle θ\thetaθ, the maximum height is

H=u2sin⁡2θ2gH = \frac{u^2 \sin^2 \theta}{2g}H=2gu2sin2θ​
  1. Projectile at 30∘30^\circ30∘
H1=u2sin⁡230∘2gH_1 = \frac{u^2 \sin^2 30^\circ}{2g}H1​=2gu2sin230∘​

Since

sin⁡30∘=12\sin 30^\circ = \frac{1}{2}sin30∘=21​

we get

H1=u2(12)22g=u28gH_1 = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g}H1​=2gu2(21​)2​=8gu2​
  1. Projectile at 60∘60^\circ60∘
H2=u2sin⁡260∘2gH_2 = \frac{u^2 \sin^2 60^\circ}{2g}H2​=2gu2sin260∘​

Since

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

we get

H2=u2(32)22g=3u28gH_2 = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3u^2}{8g}H2​=2gu2(23​​)2​=8g3u2​
  1. Ratio of maximum heights
H1:H2=u28g:3u28g=1:3H_1 : H_2 = \frac{u^2}{8g} : \frac{3u^2}{8g} = 1:3H1​:H2​=8gu2​:8g3u2​=1:3
  1. Option check
  • A: 1:31: \sqrt{3}1:3​ ❌
  • B: 3:1\sqrt{3}: 13​:1 ❌
  • C: 1:31:31:3 ✅
  • D: 2:32: \sqrt{3}2:3​ ❌

Hence, the correct answer is C.

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