https://i.stack.imgur.com/FA6xX.png
I don't understand how this is wrong. I tried AF and OC. Either of those should be right, since those depict 45 angles. What's the problem?
Asked
Active
Viewed 29 times
0
-
If the vertical acceleration is constant (and negative), how does the vertical velocity change? – Fabio Somenzi Feb 20 '17 at 04:33
-
I don't understand what you mean. – user416503 Feb 20 '17 at 04:35
-
If you read the detailed solutions that were posted in the meanwhile, you should be able to understand. – Fabio Somenzi Feb 20 '17 at 04:37
2 Answers
1
You are not asked the position, you are asked about the velocity in the vertical direction. The horizontal axis of the graph is time. Since only gravity is acting in the vertical direction $v_y(t)=v_y(0)-gt$. Therefore the graph is linear in time, with a negative slope. So the answer is AE
Andrei
- 37,370
-
I dodn't understand. It says above the horizontal line. Shouldn't that mean it's angled upwards. – user416503 Feb 20 '17 at 04:39
-
The position is going up and down, but the velocity starts positive, then goes negative. you are talking about different things – Andrei Feb 20 '17 at 04:41
1
Assuming $v_y$ means the vertical velocity, then it starts off positive (the cannon is shooting upwards), and decreases with time. Assuming the acceleration is constant (given by gravity), the velocity is an affine function of time. The only option that fits these is $AE$.
Fimpellizzeri
- 23,126