practice/lab
directions: answer the following questions based on our notes.
part i - conceptual
a student throws a ball straight upward and it takes 3 seconds to reach the top. answer the following questions.
question
1. what is the velocity of the ball at the very top?
explain why!
2. what is the acceleration of the ball going up?
explain why!
3. is the acceleration of the ball going down the same or different as the acceleration going up?
explain why!
4. is the ball faster when the student throws the ball or right before the ball hits his hand coming back down? explain why!
5. does it take longer for the ball to be thrown upward or come back down?
explain why!

Question

Accepted Answer

### Question 3
# Brief Explanations:
When the ball moves up or down (ignoring air resistance), the only force acting on it is gravity. So the acceleration is always $ g = 9.8\,\text{m/s}^2 $ downward (or $ - 9.8\,\text{m/s}^2 $ if upward is positive). Thus, the acceleration going down is the same as going up.
# Answer:
The acceleration of the ball going down is the same as going up. Because the only force acting on the ball (ignoring air resistance) is gravity, so the acceleration is always $ 9.8\,\text{m/s}^2 $ downward (or $ - 9.8\,\text{m/s}^2 $ for upward - positive coordinate system).

### Question 4
# Brief Explanations:
By the principle of conservation of mechanical energy (ignoring air resistance), the kinetic energy when thrown and when it comes back to the same height (the student's hand) is the same. Since $ KE=\frac{1}{2}mv^{2} $ and mass $ m $ is constant, the speed $ v $ is the same.
# Answer:
The ball has the same speed when the student throws it and right before it hits his hand coming back down. Because (ignoring air resistance) mechanical energy is conserved. At the same height (the hand's height), the gravitational potential energy is the same, so the kinetic energy (and thus speed, since mass is constant) is the same.

### Question 5
# Brief Explanations:
Ignoring air resistance, the time to go up and the time to come down are equal. The upward motion is a decelerated motion (final velocity $ v = 0 $ at the top) with $ v = v_0 - gt_{\text{up}} $, and the downward motion is a free - fall from rest (initial velocity $ u = 0 $) from the same height $ h=\frac{1}{2}gt_{\text{down}}^{2} $ (also $ h = v_0t_{\text{up}}-\frac{1}{2}gt_{\text{up}}^{2} $ and $ v_0=gt_{\text{up}} $ from upward motion). Solving these, we get $ t_{\text{up}} = t_{\text{down}} $. If we consider air resistance, the time to come down is slightly longer, but in the ideal case (no air resistance), they are equal.
# Answer:
In the ideal case (ignoring air resistance), it takes the same amount of time for the ball to be thrown upward and to come back down. Because the upward deceleration (due to gravity) and downward acceleration (due to gravity) are the same in magnitude, and the displacement (up and down) is the same. If air resistance is considered, coming down takes slightly longer, but typically in basic physics problems, we assume no air resistance, so times are equal.

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Question 3

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Question 4