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Motion in A Plane question

2022 · 27 Jun · Shift 1 · Q42
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  5. /2022 · 27 Jun · Shift 1 · Q42

Motion in A Plane question

2022 · 27 Jun · Shift 1 · Q42

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A projectile is launched at an angle 'α\alphaα' with the horizontal with a velocity 20 ms −-− 1. After 10 s, its inclination with horizontal is 'β\betaβ'. The value of tan β\betaβ will be : (g = 10 ms −-− 2).
  1. A
    tan α\alphaα + 5sec α\alphaα
  2. B
    tan α−\alpha-α− 5sec α\alphaα
  3. C
    2tan α−\alpha-α− 5sec α\alphaα
  4. D
    2tan α+\alpha+α+ 5sec α\alphaα
View written solutionFree

Correct answer: B

  1. Resolve the initial velocity into components

Given launch speed u=20 m s−1u=20\,\text{m s}^{-1}u=20m s−1 at angle α\alphaα:

ux=20cos⁡α,uy=20sin⁡αu_x = 20\cos\alpha, \qquad u_y = 20\sin\alphaux​=20cosα,uy​=20sinα

  1. Velocity components after 10 10\,10s
  • Horizontal velocity remains constant:

vx=20cos⁡αv_x = 20\cos\alphavx​=20cosα

  • Vertical velocity after time ttt is:

vy=20sin⁡α−gtv_y = 20\sin\alpha - gtvy​=20sinα−gt

With g=10 m s−2g=10\,\text{m s}^{-2}g=10m s−2 and t=10 t=10\,t=10s,

vy=20sin⁡α−10×10=20sin⁡α−100v_y = 20\sin\alpha - 10\times 10 = 20\sin\alpha - 100vy​=20sinα−10×10=20sinα−100

  1. Inclination of velocity with the horizontal after 10 10\,10s

If the inclination is β\betaβ, then

tan⁡β=vyvx\tan\beta = \frac{v_y}{v_x}tanβ=vx​vy​​

So,

tan⁡β=20sin⁡α−10020cos⁡α\tan\beta = \frac{20\sin\alpha - 100}{20\cos\alpha}tanβ=20cosα20sinα−100​

Simplify:

tan⁡β=20sin⁡α20cos⁡α−10020cos⁡α\tan\beta = \frac{20\sin\alpha}{20\cos\alpha} - \frac{100}{20\cos\alpha}tanβ=20cosα20sinα​−20cosα100​

tan⁡β=tan⁡α−5cos⁡α\tan\beta = \tan\alpha - \frac{5}{\cos\alpha}tanβ=tanα−cosα5​

tan⁡β=tan⁡α−5sec⁡α\tan\beta = \tan\alpha - 5\sec\alphatanβ=tanα−5secα

  1. Match with options

This corresponds to:

Option B:

tan⁡α−5sec⁡α\boxed{\tan\alpha - 5\sec\alpha}tanα−5secα​

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