Answer:
it tells you that the speed increases until about 20 seconds then keeps a steady pace for 20 seconds then the speed drops and stops at 55 seconds in the process.
Freezing (liquid to solid)
Deposition (gas to solid)
Condensation (gas to liquid)
All three of these state changes are a result of a energy loss. When considering energy loss it is best to think of situations where temperature has dropped. Less energy in the system results in less energy the substance is exposed to or has available.
16,000 m/s
Since it’s speed, and the distance is unknown. Gravity isn’t applying a noticeable force too on the rocket, as if it were, then the rocket would be accelerating negatively.
In the vertical direction, we know that acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity), we know our initial velocity (vi) = 15m/s and our final velocity (vf) = 0m/s if the ball stops at its max height. Using this kinematics equation: vf^2 = vi^2 + 2a*h, where h is the max height of the ball, we can find h by solving: h = (vf^2 - vi^2)/2a, now we plug in numbers h = (0^2 - 15^2)/2(-9.8) = 11.48 m.
Now we can find the time, but let's use a trick, we know the acceleration will be the same going up and down and that it will cover the same distance. Using this logic, we can know that the ball will be going the exact same speed we threw it up at when it comes back down to our hand (provided our hand is at the same height). We can then say it will take the same amount of time to reach its peak after leaving our hand as it will take to go from its peak back down to our hand. Using this, we can just get the time it takes to get to the top of its arc and then multiple it by 2. Using d = (vf + vi)*t / 2, we can solve for t, so t = 2d/(vf+vi) = 2(11.48)/(15) = 1.53 s for the trip up, doubling it for the trip down, we get a total of 3.06 s to go up and back down to our hand.
