# Simple Projectile Motion Help

1. May 29, 2012

### jillime

Hey guys, I'm not great at physics and my brain sometimes just doesn't get the sort of "implied" aspects of a problem. So I could use some help with an easy question.

In projectile motion, there are certain constants given particular circumstances, right?

I have to find the height of a table given the speed of a ball rolling off of it, and the distance away from the table that it lands.

But I don't know how to fill in the holes. For example, I keep wondering about velocity in the Y direction. I don't know if its 9.8, or if that's the acceleration, and I just confuse myself with my lack of confidence about what's what, what's constant, zero, etc.

having not given you guys any mathematical insight into this problem, could someone just relay to me what you would already know, given this particular circumstance?

2. May 29, 2012

### dimension10

For the first part, try using s=1/2 gt^2. You can calculate t using the initial velocity and the final velocity. I'm assuming the initial and final velocities are given. So, for example, if the initial velocity is 0 and the final velocity is 9.80665m/s^2, then the time is 1s. Then s=g/2, the numbers, that is. Around 4.9033m, if you consider the significant figures.

For the second part, try to break the velocity vector down into 2 components. The y-component would be v_0+gt. Where g is acceleration due to gravity ~9.8066m/s^2. The velocity is certainly not 9.8, its the acceleration. The velocity is not constant as gravity is acting on the ball.

Sorry for not using $\LaTeX$

3. May 29, 2012

### Villyer

How I approach this is taking what you know and asking myself "What can I find out with this?".

You know the initial horizontal velocity, and horizontal displacement. By realizing that the horizontal acceleration is 0, you can find the time.
Now, in the vertical direction, you know time (since time is the same for each direction in projectile motion). You also know that the ball is rolling off of a table, so its initial velocity in the y direction is 0. You also know that the acceleration in the y direction is g. So using those three known quantities, you can find the vertical displacement.

I guess that the four bold statements are then the implied givens for this problem.

4. May 30, 2012

### azizlwl

This a 2 dimension problem.

Normally we encountered 1 dimension case if a car is moving along a straight line. It will be a distance from the origin by multiplying velocity with time. If it is accelerating we use SUVAT equation.

In the 2 dimension, x or y are lines.
In 3D, x, y and z are planes.

Now back 2D, displacement x is a vertical line from +∞ to -∞ intersect x-axis at x unit from origin. Likewise the y line.

Both lines are moving according its function, for x=vt
and for y=ut+1/2at2
.

Thus we are only interested in the intersection on both lines, since there is only one object.
The intersection is the point where the object lies