- #1
Alboin
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Homework Statement
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
Homework Equations
(Nothing really specific)
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