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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
 one year ago
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
 one year ago

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Mimi_x3Best ResponseYou've already chosen the best response.0
well how do i start? anyone able to give me a hint?
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.3
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

Mimi_x3Best ResponseYou've already chosen the best response.0
ok i got it. how about a?
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.0
for b its the touchdown velocity right?
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.3
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

Ishaan94Best ResponseYou've already chosen the best response.3
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
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