A mass is dropped from a height of 30m, and at the same time from ground level (and directly beneath the dropped mass), and another mass is thrown straight up with a speed of 15 m/s.
a) How far apart are the two masses after 0.5 s?
b) Will the massses ever meet or pass each other? And if so will the meeting place take place while both masses are falling or when one is falling and the other is rising?
a) I got 30 - 15t
30-15(.5 seconds)=22.5 meters
b) I need help on this part of the queston.
distance apart= 30- 1/2 g (t)^2 - 15t + 1/2 g t^2
distance apart= 30 -15t
t= .5 second
If they meet, distance apart is zero
solve for t.
See what the velocity of the ball thrown upwards is at that time. Is it positive (upward) or not.
I got t= 2 seconds so would they meet when one is rising and the other is falling?
You need to determinse how long it takes for the object thrown upward to reach a top. When you do that you'll know what part of the path the object is in, rising or falling.
You know the falling ball is going down. The question is what is the rising ball doing at time = 2 seconds.
v = initialvelocity +g *time
= 15 - 9.8 *t
a positive answer is going up, a negative answer is going down.
so I would get -4.6 as the answer if I plugged in t=2?
No, look at the equation BobP. gave and use 2 in that equation as he said. Thus
v = initialvelocity +g *time or,
v = 15 - 9.8 *t
Setting t=2 we have
v = 15 - 9.8*2 = -4.6
The negative sign tells us that the object thrown upward is now falling. Therefore the two objects meet (i.e., are at the same height) when the second object is on the way down. Therefore both are falling for b).
In short, after you use the formula for v Bob gave there are no other calculations to do.
After looking at your question again I see that I may've misunderstood what you were asking.
Yes! you should get -4.6 when you put t=2 in that formula. The only thing we were interested in though was the sign of the result -4.6. since it's negative, that means it's falling. It takes approximately 1.53 sec for the thrown object to reach it's peak and then begin to fall back down. Therefore it's on it's way down when the dropped object reaches it.
So yes, you did the correct calculation. All we we're interested to know was whether it was ascending or descending and Bob gave the formula for that.
To summarize:
a) The two masses are initially 30 meters apart. After 0.5 seconds, the mass thrown upwards will have reached a height of 15 meters (since it is moving against gravity at a constant speed). Therefore, the distance apart is 30 - 15 = 15 meters.
b) To determine if the masses will meet, we need to calculate the time at which they will be at the same height. We can set the distance apart equation equal to zero and solve for time:
distance apart = 30 - 15t
0 = 30 - 15t
15t = 30
t = 2 seconds
Next, we need to determine the velocity of the thrown mass at t = 2 seconds. Using the equation v = initial velocity + g * time, we get:
v = 15 - 9.8 * 2
v = -4.6
Since the velocity is negative, it means the mass is descending. Therefore, the two masses will meet when the thrown mass is descending (falling back down) and the dropped mass is also falling.
In conclusion, the masses will meet when one is falling and the other is also falling.