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ollegr [7]
3 years ago
15

(This is a non-relativistic warm-up problem, to get you to think about reference frames.) A girl throws a baseball upwards at ti

me t=0. She catches it at exactly t=2.0 seconds. A) Calculate the position of the ball as a function of time (0
Physics
1 answer:
____ [38]3 years ago
5 0

Answer:

X(t) = 9.8 *t - 4.9 * t^2

Explanation:

We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.

We use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a *t^2

X0 = 0 because it is at the origin of the coordinate system.

We know that at t = 2, the position will be zero.

X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2

0 = 2 * V0 - 4.9 * 4

2 * V0 = 19.6

V0 = 9.8 m/s

Then the position of the ball as a function of time is:

X(t) = 9.8 *t - 4.9 * t^2

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Suppose you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six months, as measured by
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t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2} } }\\

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Therefore,

90\ yr = \frac{0.5\ yr}{\sqrt{1-\frac{v^2}{c^2} } } \\\\\\\sqrt{1-\frac{v^2}{c^2}} =  \frac{0.5\ yr}{90\ yr}\\ 1-\frac{v^2}{c^2} = 0.00003086\\\frac{v^2}{c^2} = 1-0.00003086\\

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