# Homework Help: Help setting up problems

1. Feb 7, 2004

### Pepsi24chevy

Hi, i was wondering if you could help direct me in setting up these problems.

1, A ball is thrown upward with an initial velocity of 20m/s. A) how long is the ball in the air? B0 What is the greatest height reached by the ball? C0 How long after release is the ball 15m abovet he release point?

2. 80. An objecat is released from rest at a height h. DUring the final second of its fall, it traverses a distance of 38m. What is H?

3. A particle moves with a constant acceleration of 3m/s^2. At a time of 4 s, it is at a position of 100m with respect to soem coordinate system; at a time of 6s, it has a velocity of 15m/s. Find its position at a time of 6s.

So any help in setting up these problems would be appreciated. Thanks

2. Feb 7, 2004

### jamesrc

These problems deal with 1-D kinematics of particles under constant acceleration. The equations that end up being the most useful in these types of problems are:

$$v = v_0 + at$$
$$v^2 = v_0^2 + 2a\Delta x$$
$$x = x_0 + v_0t + \frac{at^2}{2}$$

3. Feb 8, 2004

### Pepsi24chevy

FOr the second problem, i used, Vf^2= Vi^2+2*a*d and i get 27.29, but this does not seem right since it traverses a distrance of 38m in the last second. None of the other equations seem to fit.

For the third problem, am i going to want to find the final velocity of the particle at 4s and then how would i use this info to find out the positon of the particle at 6 seconds. For the 3rd problem i did 100+(15*6)+(3*6^2)/2 and got 244. I dunno if this is the answer or what. My teacher isnt' the best and got rated as one of the worst teachers for explaining stuff, but he is only teacher for physics course i have to take.

Last edited: Feb 8, 2004
4. Feb 8, 2004

### jamesrc

For the second problem, I would use:

$$\Delta y_b = v_{oy}t + \frac{at^2}{2}$$

and solve for the initial velocity of this portion of the fall.
Make sure you keep your signs straight here. &Delta;yb = -38 m. (the b stands for bottom), a = -9.81 m/s/s, and voy should turn out to be negative.

Then I would use

$$v_{oy}^2 - v_o^2 = 2a\Delta y_t$$

Solve for &Delta;yt to find

$$h = |\Delta y_t| + |\Delta y_b|$$

Notice that the "initial" velocity from the first equation is now the "final" velocity in this second equation.

For the third problem, since you know that the velocity at t = 6 s is +15 m/s and you know that the acceleration a = +3m/s/s and is constant, you can find the velocity at t = 4s. (use v = vo + a&Delta;t). You can then use the following to find the final position of the object:

$$x = x_o + v_o(\Delta t) + \frac{a(\Delta t)^2}{2}$$

&Delta;t in both equations is 6-4 = 2 seconds, xo = 100 m.

I hope this has cleared things up a bit.

5. Feb 8, 2004

### Pepsi24chevy

Ok for the second one, would the answer be 88.92? And for the 3rd, i get 106 or 142. WOuld the inital velocity for the first part of 3 be 0?

6. Feb 8, 2004

### holly

Pepsi, I have similar problems but not as hard as yours. But try drawing them out, drawing out the parabola for the tossed-upward ball, for example, and remembering that it's going to be symmetrical, time up equalling time down (assuming freefall). Plus, draw out a ball falling down off a cliff (assuming freefall), go ahead and write down where it is at one second, at two seconds, how far it has gone at each second, etc. Even the real physicists often draw things out, and it helped me. Like at one second, it has gone approx 5 meters, one second has passed, and the accel is gt**2 (we use 10 for g)...if you have these on an exam, first thing you do is draw it all out on the back of the test or scrap paper.

7. Feb 8, 2004

### jamesrc

For problem 2, I get closer to 94 m.

For the 3rd problem I get 124 m. If by the initial velocity, you mean the velocity at t = 0, that would be -3m/s, but that is irrelevant to the solution of the problem.

If you're still having trouble, please type out the work you've done and we'll try to find where the error is.

8. Feb 8, 2004

### Pepsi24chevy

for 2, i put -38=Voy + -4.905
Voy= -33.095

-33.095^2 +0^2/ 19.62= -55.825
Then i added 33.095 + 55.825 and got 88.9.

For 3, i got x= 100+ 12+ 6
and got 118.

I appreciate the help, i have been lookin up websites on teh internet and going through to try to learn this stuff.

Last edited: Feb 8, 2004
9. Feb 8, 2004

### jamesrc

Hi,

For #2, you did it right, except you should be adding 55.8 + 38.
You added a distance to a velocity, which is a no-no.

For #3, it should be

x = 100m + (9m/s)*2s + .5*(3m/s/s)*(2s)2
x = 100 + 18 + 6 = 124m

It looks like the part you missed there was the velocity at t = 4s:

v(t=6s) = 15 m/s, right (given)
a = 3m/s (also given)
so 15 = vo + 3*2 --> vo = 9 m/s

(maybe v4 would have been a more appropriate variable to use, but hopefully you get the idea.)