Question

A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.6 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.8 m/s from a height of 28.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
1) What is the speed of the blue ball when it reaches its max. height?
2)How long does it take the blue ball to reach its maximum height?
3)What is the maximum height the blue ball reaches?

Answer 1

4)What is the height of the red ball 3.64 seconds after the blue ball is thrown?
5)How long after the blue ball is thrown are the two balls in the air at the same height?

I got the second answer but I just can't seem to get any of the others... I've been trying for hours! Please help!

Answer 2

1) Think about what is happening to the blue ball as it approaches its max height.

2) You can use \[\frac{v-u}{t}=a\]3) You can use \[s=vt\], but don't forget to add on the 0.6 metres of height that it started off with!

Answer 3

4) It's an oddly phrased question, but that's 0.84 seconds after the red ball is dropped. Use \[\frac{v-u}{t}=a\]5) This is a tough one, you could probably say that the height of the blue ball is given by \[y_{blue}(t)= 0.6 + v_{blue}t + \frac{1}{2}at^2 \text{, and the height of the red ball is}\]\[\text{ given by }y_{red}(t)=28.9 + v_{red}t + \frac{1}{2}at_2\]Then you're looking for the place there they are equal to each other.

As a side note, the equation given for 3) is incorrect, it should be \(y_{blue}(t)= 0.6 + v_{blue}t + \frac{1}{2}at^2\)

Answer 4

Also, take care that your velocities have the right signs (take positive as being upwards, negative as being downwards)

Answer 5

Ah sounds tough but I'm sure that will help me a ton, thanks alot!

Answer 6

If you have troubles, just post 'em!

Answer 7

I managed to get the right answers and understand everything except number 5. If I plug everything in for the blue ball for example wont I just get the blue ball at max height, this question is just very confusing to me.

Answer 8

You will yes. Perhaps I didn't explain it well. You're looking for the point where \[y_{blue}(t)=y_{red}t\]so you let those two equations equal each other, and solve for t.

I think I made yet another mistake, by forgetting to take into account the time the red ball starts falling. I think it should be \[y_{red}(t) = 28.9 + v_{red}(t-2.8) + \frac{1}{2}a(t-2.8)^2\]

Answer 9

Aha I understand that then thanks!

Answer 10

Ah I dont mean to bother but I used that exact formula using vblue=23m/s, a=9.81m/s^2... I think whats confusing me is vred because earlier I found it to be 15.0404 at a certain time, I dont have a time so I dont exactly know what vred is, as I dont think I should be using initial velocity for it. Argh so confusing!

Answer 11

I think I'm going to have to give up on this last question, thanks for all the help though!