Answer:
Question 1: <u>1 s after the motion starts</u>
Question 2: <u>0 (just when the motion starts)</u>
Explanation:
You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.
<u>1. From the speedometer shown at the right.</u>
You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.
- Free fall equation: Vf = Vo - gt
- Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g
For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/
- t = [10 m/s] / [10 m/s²] = 1 s
That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.
<u>2. Time when the speedometer displays a reading of 20 m/s</u>
First, calculate the launching speed:
Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:
- Vo = gt = 10 m/s² × 3 s = 30 m/s
Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s
- 20 m/s = 30 m/s - 10 m/s² × t
- t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s
Thus, the first answer is t = 1 s.
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<u>3. Time when the speedometer displays a reading of 30 m/s</u>
This is the same speec estimated for the launching: 30 m/s.
So, this reading corresponds to the moment when the ball was launched.
Thus time is 0, i.e. it is the same instant of the launch.
If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.
That is why I recommended to work with g = 10 m/s².
