Answer: i think that the third option is correct
the force of the ball goes to the floor making it go down and up reflecting a little lower force back to the ball and sending the ball back up
hope this helps
brainliest if possible
<span>v/2
This is an exercise in the conservation of momentum.
The collision specified is a non-elastic collision since the railroad cars didn't bounce away from each other. For the equations, I'll use the following variables.
r1 = momentum of railroad car 1
r2 = momentum of railroad car 2
x = velocity after collision
Prior to the collision, the momentum of the system was
r1 + r2
mv + m*0
So the total momentum is mv
After the collision, both cars move at the same velocity since it was non-elastic, so
r1 + r2
mx + mx
x(m + m)
x(2m)
And since the momentum has to match, we can set the equations equal to each other, so:
x(2m) = mv
x(2) = v
x = v/2
Therefore the speed immediately after collision was v/2</span>
To solve this problem we use kinematics formulas.
to. What was the acceleration of the rocket during the first 16-s?
![v_1 = v_o + a_1t_1](https://tex.z-dn.net/?f=v_1%20%3D%20v_o%20%2B%20a_1t_1)
Where
= speed after 16 s
= initial velocity = 0
= acceleration during 16 s
![v_1 = v_0 + a_1t_1](https://tex.z-dn.net/?f=v_1%20%3D%20v_0%20%2B%20a_1t_1)
![v_1 = 0 + 16a_1](https://tex.z-dn.net/?f=v_1%20%3D%200%20%2B%2016a_1)
![v_1 = 16a_1](https://tex.z-dn.net/?f=v_1%20%3D%2016a_1)
Now we use the formula for the position:
![h_1 = h_0 + v_0t_1 + 0.5a_1t_1 ^ 2](https://tex.z-dn.net/?f=h_1%20%3D%20h_0%20%2B%20v_0t_1%20%2B%200.5a_1t_1%20%5E%202)
Where:
= position after 16 s
= initial position = 0
= 16 s
![h_1 = 0 + 0 + 0.5a_1(16)^2](https://tex.z-dn.net/?f=h_1%20%3D%200%20%2B%200%20%2B%200.5a_1%2816%29%5E2)
![h_1 = 128a_1](https://tex.z-dn.net/?f=h_1%20%3D%20128a_1)
Then, we know that the altitude of the rocket after 20 s is 5100 meters.
Then we will raise the equation of the position of the rocket from the instant
to ![t_2](https://tex.z-dn.net/?f=t_2)
![t = t_2 - t_1\\t = 20 - 16\\t = 4\ s](https://tex.z-dn.net/?f=t%20%3D%20t_2%20-%20t_1%5C%5Ct%20%3D%2020%20-%2016%5C%5Ct%20%3D%204%5C%20s)
where:
![h_2 = h_1 + v_1t - 0.5gt^2](https://tex.z-dn.net/?f=h_2%20%3D%20h_1%20%2B%20v_1t%20-%200.5gt%5E2)
![h_2 = 128a_1 + 16a_1t - 0.5(9.8)t^2](https://tex.z-dn.net/?f=h_2%20%3D%20128a_1%20%2B%2016a_1t%20-%200.5%289.8%29t%5E2)
![5100 = 128a_1 + 16a_1(4) -0.5(9.8)(4)^2](https://tex.z-dn.net/?f=5100%20%3D%20128a_1%20%2B%2016a_1%284%29%20-0.5%289.8%29%284%29%5E2)
Now we clear
.
![5100 +78.4 = a_1(128 + 64)](https://tex.z-dn.net/?f=5100%20%2B78.4%20%3D%20a_1%28128%20%2B%2064%29)
![a_1 = 26.97 m/s^2](https://tex.z-dn.net/?f=a_1%20%3D%2026.97%20m%2Fs%5E2)
The aceleration is ![26.97 m/s^2](https://tex.z-dn.net/?f=26.97%20m%2Fs%5E2)
So:
![v_1 = a_1t\\\\v_1 = 16(26.97)\\\\v_1 = 431.52 m/s](https://tex.z-dn.net/?f=v_1%20%3D%20a_1t%5C%5C%5C%5Cv_1%20%3D%2016%2826.97%29%5C%5C%5C%5Cv_1%20%3D%20431.52%20m%2Fs)
What is the speed of the rocket when it passes through a cloud at 5100 m above the ground?
![v_2 = v_1 - gt\\\\v_2 = 431.52 - 9.8(4)\\\\v_2 = 392.32 m/s](https://tex.z-dn.net/?f=v_2%20%3D%20v_1%20-%20gt%5C%5C%5C%5Cv_2%20%3D%20431.52%20-%209.8%284%29%5C%5C%5C%5Cv_2%20%3D%20392.32%20m%2Fs)
The speed is 392.32 m/s