## Mimi_x3 3 years 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

1. Mimi_x3

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2. Mimi_x3

well how do i start? anyone able to give me a hint?

3. Vaidehi09

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.

4. Mimi_x3

ok i got it. how about a?

5. Mimi_x3

for b its the touchdown velocity right?

6. Vaidehi09

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.

7. Ishaan94

Range is same. $\frac{u^2 \sin2\alpha}{g} = \frac{u^2\sin 2\beta }{g}\implies \sin 2\alpha = \sin2\beta$ $\implies 2\alpha = 2\beta\;or\; \pi - 2\alpha = 2\beta \implies \frac{\pi}2 = \alpha +\beta$

8. Mimi_x3

thank you!