How do I calculate the maximum vertical displacement of a cart as it reaches the open end of a track going 20 m/s at an angle of 30 degrees?
- Thread starter Squablo
- Start date
Maths
You take the square root of pie, multiply it by the number of people in your park, divide it by the current weather and then simply add 6.3. This only works on a Tuesday though
Projectiles. First of all you want to work out the starting velocity u, in the x and y directions. To do this you used trigonometry, to find the opposite and adjacent sides. You set the ending vertical velocity to 0ms^-1 (because you know this is 0 at maximum vertical displacement, until it later falls to the ground under gravity), and the horizontal acceleration is presumed 0 (assumption that there is no air resistance). Your acceleration vertically is due to gravity, acting against the vertical direction of the cart (-9.81ms^-2). This cart is modelled presumably as a point mass(?) with no added weight, so we can use the kinematic equation of motion (v^2-U^2)/2g = s. Where s is the maximum vertical displacement. I get 0.51m
Since I am not sadistic ([down]) I hope they won't crash.