Help with homework problems please

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 3K views
PreDeadMan123
Messages
4
Reaction score
0
Help with homework problems please!

A cannon is mounted 50m away from the edge of a cliff that drops 10m to the plain below. If the cannon projecticle is fired with an initial velocity of 75 m/sec find the angle of elevation so that the shell lands closest to the cliff face from the plane below.
 
Physics news on Phys.org
Welcome to PF!

Please show some working and / or intial thoughts. In addition we have a section dedicated to homework or coursework questions.

~H

Note to mentors: please move to Intro Physics
 
i have no clue i am new to physics this is hard as heck so many formulas but.. ah could someone explain to me how to do this I'm confused
 
Basically, you can split the velocity into two components, one vertical (in the y-direction), and one horizontal (in the x-direction). The projectile is fired at an angle, therefore we need to resolve the velocity vector into it's x and y components. If we let [itex]\theta[/itex] be the angle of elevation above the horizontal the x and y components are;

[tex]v_{x} = 75\cos\theta[/tex]

[tex]v_{y} = 75\sin\theta[/tex]

Vertical motion is independent of horizontal motion. Using kinematic equations can you find an expression for the time of flight of the projectile? HINT: Just consider vertical motion.

~H
 
ah I am just to confused which formula do i use?
 
PreDeadMan123 said:
i have no clue i am new to physics this is hard as heck so many formulas but.. ah could someone explain to me how to do this I'm confused

Itis very very difficult to cover stuff from the ground up on such a forum. We can not replace your teacher or textbook. There is only so much that can be done here. The best way to use the homework section is to learn most of the material in class or in a book in order to learn and understand the fundamentals. Then, if you are stuck on a specific issue or calculation, people can help efficiently.