Projectile motion, need to find angle

by frosti
Tags: angle, motion, projectile
frosti is offline
Oct14-08, 01:28 PM
P: 13
1. The problem statement, all variables and given/known data
A cannon with a muzzle speed of 1009 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 795 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)

2. Relevant equations
I'm thinking this equation is my best bet, yf = tan(angle)xf - (g/(2 * vi^2 * cos^2(angle))) * (xf^2)

3. The attempt at a solution
since I know the vi, xf, and yf, I was able to plug everything in and reached 795 = 1900tan(angle) - 17.375/cos^2(angle). This is where I am stuck. I tried several times but could not find a way to solve for angle.

I would really appreciate it if someone could point out how to solve for the angle in my problem. Or is there a totally different way of solving this problem?
Phys.Org News Partner Science news on
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
physics girl phd
physics girl phd is offline
Oct14-08, 02:24 PM
physics girl phd's Avatar
P: 938
Your work DOES look great (I checked your first equation). The problem is that you've got an algebraically difficult thing to solve for (the angle) because it's inside trig functions.

What I suggest is keeping your equation in terms of the components. Relate one of these components ONLY to the initial speed and the remaining component. Then solve for the component you kept. You can use this solution to find the angle.

Added at a later edit: You could also just graph your second function and look for where it crosses zero, but that's easy only because of modern graphing tools.
frosti is offline
Oct14-08, 06:26 PM
P: 13
thank you for your input, i separated my problem into x and y components:
x component ---> 1900 = 1009cos(theta) * t
y component ---> 795 = 1009sin(theta) * t - 4.9t^2

I used my x component to solve for cosine
1900/1009 = cos(theta) * t
cos(theta) = 1.883/t

I then used trig identity sin^2(theta) + cos^2(theta) = 1
sin^2(theta) + (1.883/t)^2 = 1
sin(theta) = square root(1 - (1.883/t)^2)

then I plugged sin(theta) equation into my y component
795 = 1009 * square root(1-(1.883/t)^2) * t -4.9t^2

I think I'm on the right track but I don't really know where to go with this, it's not really a quadratic equation and I don't know what to do now. Any pointer would be greatly appreciated.

frosti is offline
Oct14-08, 06:57 PM
P: 13

Projectile motion, need to find angle

nm, I got it. Thank you for all your help.

Register to reply

Related Discussions
Projectile Motion given angle and distance, find intial velocity Introductory Physics Homework 5
[SOLVED] Projectile Motion - Ramp Angle Introductory Physics Homework 5
Find angle (Projectile Motions) Introductory Physics Homework 7
projectile motion - no angle given??? Introductory Physics Homework 2
projectile motion at an angle Introductory Physics Homework 2