Homework Help: Simple Linear Motion Question

1. Sep 23, 2009

Alboin

1. The problem statement, all variables and given/known data
A ball is thrown vertically into the air and is caught 3.1 seconds later. (This is just a basic linear motion question, so there are no 'advanced' considerations.)

a) Find the velocity when the ball was thrown
b) Find the max height

2. Relevant equations
(Nothing really specific)

3. The attempt at a solution
I think I have this right, but most other people in my class have a completely different answer. Can you tell me what, exactly, I am missing? (Or, if I'm actually correct?)

Let a = acceleration in m\s^2, V = initial velocity in m\s, and t = time in sec. In this problem, a is -g, or -9.8.

a) The falling ball should be able to be modeled with the equation y=V*t + .5(a)(t^2). However, while the initial velocity is not known, the x of the vertex is (The parabola stretches 3.1sec. and thus its center is 3.1/2), so by using the vertex form of a quadratic:

y = -4.9(x-1.55)^2 + k

And, since we are interested in is the initial velocity, or the first derivative at 0, the k is irrelevant, and:

y' = -9.8x+15.19

Thus, at 0, the velocity is 15.19 m\s.

b) The height using the original y=V*t + .5(a)(t^2):

y = 15.19(1.55) + .5(-9.8)(.1.55)^2
= 11.8 m

What am I missing?
Thanks,
Alboin

2. Sep 23, 2009

Furby

As I understand it, in an introductory physics course, a ball thrown vertically is strictly vertical unless otherwise stated. Thus the trajectory would simply be a straight line, rather than a parabola with an x-component to the velocity.

Is there a picture or any other reason to think otherwise? Unless this is under the "2-D motion" category of your textbook or something, then you might be making this problem harder than it really is. The problem doesn't account for it, thus we could consider the drag of air on the ball, and the actual height the ball may have been caught, complicating the problem more.

3. Sep 23, 2009

Alboin

Oh, I'm sorry. y = -4.9(x-1.55)^2 + k and friends are referring to distance\height in the air (y) versus time (x). So as time increases, the distance increases, but then falls again - a parabola.

4. Sep 23, 2009

Furby

Alright, if the ball is not strictly vertical in motion, then your velocity for part a is nearly correct. You failed to account for the velocity's factor that assists in the horizontal motion of the ball, creating the parabolic trajectory.

You found v0y, not the entire resulting velocity. Your maximum height is correct, but I think you found it accidentally. You wouldn't plug in the actual initial velocity of the ball into the equation, as you tried to do.

EDIT: I think I just understood your previous post. If the trajectory isn't an ACTUAL parabola (sorry!), then yes your answer is correct, lol.