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Mimi_x3 Group Title

Projectile Motion: Two particles are projected simultaneously from \(A\) with the same velocity \(V\) at the angles of elevation \(\alpha\), \(\beta\) on a horizontal plane respectively. If they both pass through \(B\), one after time \(t_1\) and the other after time \(t_2\). (a) show that \(\alpha+ \beta\=\frac{\pi}{2}\) (b) Prove that \(4V^2 = g^2 (t_{1}^{2} + t_{2}^{2}\) Will draw diagram below

  • one year ago
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  1. Mimi_x3 Group Title
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    |dw:1344001551890:dw|

    • one year ago
  2. Mimi_x3 Group Title
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    well how do i start? anyone able to give me a hint?

    • one year ago
  3. Vaidehi09 Group Title
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    proving part b is simple. use the formula for time of flight of a projectile: T = 2vsinx/g......for both the projectiles. so u'll have t1 = 2vsinA/g.........[A = alpha]............(i) t2 = 2vsinB/g.........[B = beta]..............(ii) now, the RHS of ur question has t1^2 + t2^2..........so do exactly that. first square both (i) and (ii) and then add those. and then simplify and arrange the terms as required in the question.

    • one year ago
  4. Mimi_x3 Group Title
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    ok i got it. how about a?

    • one year ago
  5. Mimi_x3 Group Title
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    for b its the touchdown velocity right?

    • one year ago
  6. Vaidehi09 Group Title
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    still working on that one..i just remember that if the range is same for 2 projectiles then their angles of projection are complementary. i don't remember the proof though. so trying to derive that. and for b, its the initial velocity....given as V.

    • one year ago
  7. Ishaan94 Group Title
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    Range is same. \[\frac{u^2 \sin2\alpha}{g} = \frac{u^2\sin 2\beta }{g}\implies \sin 2\alpha = \sin2\beta\] \[\implies 2\alpha = 2\beta\;or\; \pi - 2\alpha = 2\beta \implies \frac{\pi}2 = \alpha +\beta \]

    • one year ago
  8. Mimi_x3 Group Title
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    thank you!

    • one year ago
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