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) solve the distance equation for each particle. I suggest this form..
y=Yo + Vo time + 1/2 a time^2
Use Positive quantities for up, and negative for down.
I will be happy to critique your thinking.
y=Yo + Vo time + 1/2 a time^2
0 = 30+0t =1/2 (-9.8)t^2
0=30 + 1/2 (-9.8)t^2
t^2 = -30/(-9.8* 2)
t=1.237 seconds
y=Yo + Vo time + 1/2 a time^2
0 =0 + (15m/s)t +1/2 (-9.8)t^2
t=3.06122 seconds
Did I do this correctly?
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)
y=Yo + Vo time + 1/2 a time^2
0 = 30+0t =1/2 (-9.8)t^2
0=30 + 1/2 (-9.8)t^2
t^2 = -30/(-9.8* 2)
t=1.237 seconds
y=Yo + Vo time + 1/2 a time^2
0 =0 + (15m/s)t +1/2 (-9.8)t^2
t=3.06122 seconds
Did I do this correctly?
Then how will i figure out how far apart the two masses are after .5 seconds? Also How can I figure out part b?
To figure out how far apart the two masses are after 0.5 seconds, we need to find the positions of each mass at that time.
For the mass that was dropped, we can use the equation:
y = Yo + Vo * t + 1/2 * a * t^2
Plugging in the values, we get:
y = 30 + 0 * 0.5 + 1/2 * (-9.8) * 0.5^2
y = 30 + 0 - 1.225
y ≈ 28.775 meters
So, after 0.5 seconds, the dropped mass is about 28.775 meters above the ground.
For the mass that was thrown upwards, we can use the same equation:
y = Yo + Vo * t + 1/2 * a * t^2
Plugging in the values, we get:
y = 0 + 15 * 0.5 + 1/2 * (-9.8) * 0.5^2
y = 0 + 7.5 - 1.225
y ≈ 6.275 meters
So, after 0.5 seconds, the mass thrown upwards is about 6.275 meters above the ground.
To find the distance between the two masses, we subtract the position of the mass thrown upwards from the position of the dropped mass:
Distance = |28.775 - 6.275|
Distance ≈ 22.5 meters
Therefore, the two masses are approximately 22.5 meters apart after 0.5 seconds.
Now, let's move on to part b. To determine if the masses will meet or pass each other, we need to compare their heights at any given time.
In this case, the mass thrown upwards will reach its highest point and start falling back down, while the dropped mass will continue to fall. Since the mass thrown upwards reaches its highest point at 3.06122 seconds and starts falling back down, we can conclude that the two masses will meet at some point.
The meeting place will take place while one mass is falling and the other is rising. Specifically, the mass thrown upwards will be in the process of falling back down, and the dropped mass will still be falling.