Go back
Projectile Motion

Projectile Motion

Posers and Puzzles

Vote Up
Vote Down

y = x*tan(theta) - (g*x^2) / (2*cos(theta)^2*v^2)

Using a launch speed (v) of 200 ms-1 find the maximum distance (x) at which a height (y) of 1000m can be achieved.

Gravity (g) can be assumed to be 9.8ms-2.

Vote Up
Vote Down

Homework?

2 edits
Vote Up
Vote Down

Originally posted by Bowmann
Homework?
You could say that.
It's an extra question and I can't get my matlab program to work properly. I believe the answer is less than but close to 3000m.

Attempts to attack the equation using Maple fail as the variables are x and theta and for some reason it doesn't seem to like that.


EDIT: Aha. Got it. 2915m with a launch angle of 0.951.

Vote Up
Vote Down

Originally posted by Bowmann
Homework?
Osama wants it handed in today 😛

Vote Up
Vote Down

Originally posted by XanthosNZ
You could say that.
It's an extra question and I can't get my matlab program to work properly. I believe the answer is less than but close to 3000m.

Attempts to attack the equation using Maple fail as the variables are x and theta and for some reason it doesn't seem to like that.


EDIT: Aha. Got it. 2915m with a launch angle of 0.951.
Glad we were able to help.