A community for students.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1well how do i start? anyone able to give me a hint?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0proving 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.

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1ok i got it. how about a?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1for b its the touchdown velocity right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0still 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Range is same. \[\frac{u^2 \sin2\alpha}{g} = \frac{u^2\sin 2\beta }{g}\implies \sin 2\alpha = \sin2\beta\] \[\implies 2\alpha = 2\beta\;or\; \pi  2\alpha = 2\beta \implies \frac{\pi}2 = \alpha +\beta \]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.