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Motion in A Plane question

2013 · Shift 0 · Q74
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Motion in A Plane question

2013 · Shift 0 · Q74

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A projectile is given an initial velocity of (i^+2j^)\left( {\widehat i + 2\widehat j} \right)(i+2j​) m/s, where i^{\widehat i}i is along the ground and j^{\widehat j}j​ is along the vertical. If g = 10 m/s2, the equation of its trajectory is:
  1. A
    y = x - 5x2
  2. B
    y = 2x - 5x2
  3. C
    4y = 2x - 5x2
  4. D
    4y = 2x - 25x2
View written solutionFree

Correct answer: B

  1. Resolve the initial velocity components

The projectile is projected with initial velocity u⃗=i^+2j^ m/s\vec u = \hat i + 2\hat j \text{ m/s}u=i^+2j^​ m/s So,

  • horizontal component: ux=1 m/su_x = 1 \text{ m/s}ux​=1 m/s
  • vertical component: uy=2 m/su_y = 2 \text{ m/s}uy​=2 m/s

Gravity acts downward with g=10 m/s2g = 10\text{ m/s}^2g=10 m/s2

  1. Write the equations of motion

For projectile motion: x=uxt=tx = u_x t = tx=ux​t=t So, t=xt = xt=x

For vertical motion: y=uyt−12gt2y = u_y t - \frac{1}{2}gt^2y=uy​t−21​gt2 Substitute the values: y=2t−12(10)t2y = 2t - \frac{1}{2}(10)t^2y=2t−21​(10)t2 y=2t−5t2y = 2t - 5t^2y=2t−5t2

  1. Eliminate time

Since t=xt = xt=x, y=2x−5x2y = 2x - 5x^2y=2x−5x2

  1. Match with the options

The trajectory is y=2x−5x2y = 2x - 5x^2y=2x−5x2 This matches Option B.

  1. Verification with stored answer

Stored correct answer: B Derived answer: B So they agree.

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