# Probably a very easy problem

1. Feb 11, 2010

### pokemonstar

http://www.whitwellhigh.com/jcantrell/cp/conceptdev/cpcd0501.pdf [Broken]

I'm looking at only the 1st page of the pdf file above

#1 How do i figure out the # of seconds the ball is in the air?
I have to draw the positions...so how do i know how many meters it is after 1 second?

for #2 should the ball be touching the ground after 4 seconds? how do i know how steep to draw the curve?

how can i explain the question to #2 (motion affected by horizontal/vertical)

I'm probably looking too much into this. Someone please explain :x

Last edited by a moderator: May 4, 2017
2. Feb 12, 2010

### CompuChip

Did you see the formula s = (1/2) a t2 before?

#2 will become much easier once you start with the last part of the question (how do the horizontal and vertical motion affect each other?). The answer, which you are going to remember for the rest of your (educational) life, is: they don't. The ball just keeps going horizontally at the same speed, no matter what the vertical motion is. So if you want to solve a physical problem, you can use two completely separate sets of equation (for uniform motion horizontally, and for uniformly accelerated motion due to gravity in the vertical direction).

Will that get you started?

3. Feb 12, 2010

### pokemonstar

first off, thank you for your response! =)

the formula my class got from my teacher was x(t) = 1/2 a $$\Delta$$t^2 + Vo$$\Delta$$t + Xo

so...i don't know. i'm a bit confused she didn't explain any of the equations she gave us thoroughly..

from what I'm getting from you. I plugged in x = 1/2 (10m/s^2) (1s) which gives me 5 m/s

since 1cm:5m I drew the ball falling down every 1 cm. I said the ball was in the air for 16 seconds, I don't know how to mathematically display that.

I got 1cn/s (1/16cm) but that's giving me 1/16th of a second.

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For #2 I drew the ball curving down at 1cm/s since you said the speed is not affected whether going straight down or being thrown. is that what you meant?

Thanks so much

4. Feb 12, 2010

### jhae2.718

The basic equation for motion with constant acceleration using time is $$x\left(t\right)=x_0+v_0t+\frac{at^2}{2}$$.

x0= the initial position
v0= the initial velocity
a = the acceleration
t = the time

This gives us the position after a certain time t. x0 is where the object starts. (v0+at/2)t gives us the displacement due to the velocity.

For two dimensional motion (without air resistance), use the parametric equations:
$$x\left(t\right)=x_0+v_{0x}t$$
$$y\left(t\right)=y_0+v_{0y}+\frac{at^2}{2}$$

5. Feb 13, 2010

### sean-820

....

Last edited by a moderator: May 4, 2017