Respuesta :
Answer:
Explanation:
Let's take this part by part. First it's asking for the speed of the red ball before it hits the ground. This one is pretty easy because it starts off in the downward direction. The formula for speed is vf^2 = vo^2 + 2ad, so we just solve this for vf.
Just mentioning I'm not including units here but you should.
vf^2 = vo^2 + 2ad
vf = sqrt(vo^2 + 2ad)
vf = sqrt((-1.2)^2 + 2(-9.8)*-25)
vf = 22.168
But remember this is in the downward direction.
2) Since we have vf there is a simple equation to find time.
vf = vo + at
(vf - vo)/a = t
(-22.168 - -1.2)/(-9.8) = t
2.14 = t
If you didn't want to use vf and only use the values given in the problem you could use d = vo*t + (1/2)at^2 instead, but that's a quadratic so a little bit less fun. It would be more accurate though since there'd be less rounding.
3) Since the blue ball is thrown upwards first, it will eventually slow down due to gravity and then be pulled down. To find the height it reaches we use vf^2 = vo^2 + 2ad again, and set vf = 0.
vf^2 = vo^2 + 2ad
(vf^2 - vo^2)/(2a)
(0 - 23.3^2)/(2*-9.8) = d
27.698 = d
4)1.7 seconds after the red ball is thrown, how long ago was the blue ball thrown? Well, the blue ball was thrown .6 seconds after the red one, so 1.7 seconds after the red ball was thrown, the blue ball was thrown 1.1 seconds ago. So we are looking for the height 1.1 seconds after the blue ball was thrown. For this we can use d = vo*t + (1/2)at^2
d = vo*t + (1/2)at^2
d = 23.3*1.1 + (1/2)(-9.8)1.1^2
d = 19.701
Or at least it would if it was being thrown from ground level. The question tells us that the blue ball is thrown .9 meters above the ground, so we add that to what we got.
19.701 + .9 = 20.601
5) This one's a bit complicated. So far you just had to pick the right equation with what information you were given. Here you need to do a good bit of algebra.
d = vo*t + (1/2)at^2 Is the best equation here. If you graph it you can actually see the path of the object, so we'll do two different graphs. First, the red ball.
d = -1.2x + (1/2)*-9.8x^2 + 25
Remember, that +25 at the end is because you are already starting throwing it 25 m above the ground. Then for the blue ball
d = 23.3(x-.6) + (1/2)*-9.8(x-.6)^2 + .9
Again, the +.9 is because you start .9 m above the ground. the x+.6 in the two parts is because in this x represents time, and you are throwing it .6 seconds after the red ball. So you are moving the equation to the right by .6. If you graph it and look at the value .6 you should get .9, representing .6 seconds after the red ball was thrown, this ball is at .9 meters, just as it is being thrown.
Now, we set the two equations equal to each other and solve for the time.
-1.2x + (1/2)*-9.8x^2 + 25 = 23.3(x-.6) + (1/2)*-9.8(x-.6)^2 + .9
Let's simplify, carrying out any multiplication or adding or subtracting we can.
-1.2x -4.9x^2 +25 = 23.3x - 13.98 -4.9(x^2 - 1.2x + .36)
Had to do some foiling on the right side so combine terms one more time.
-1.2x -4.9x^2 +25 = 23.3x - 13.98 - 4.9x^2 + 5.88x - 1.764
-1.2x -4.9x^2 +25 = 29.18x - 15.744 - 4.9x^2
Now we'll move everything to one side. So we can add 4.9x^2 to both sides to get rid of it on one, though in this case it gets rid of it on both. Do this with all terms of one side so they are on the other. I am going to take everything on the right and move it to the left.
-1.2x + 25 = 29.18x - 15.744
-30.38x +25 = -15.744
-30.38x + 39.744 = 0
So I messed up a bit and it's turned from a quadratic to a linear equation. It just has an x rather than an x^2. Still, this is fine, just solve for the x.
-30.38x + 39.744 = 0
39.744 = 30.38x
39.744/30.38 = x
1.308 = x
So there's your answer. 1.308 seconds. Double check witht he two graphs. And remember, for most problems like this you just need to know the different physics formulas, pick the one you need based on the information given and then plug in relavant information. Keeping in mind correct signs, if it's positive or negative. Hopefully this made sense, I'd be happy to answer any questions though.
