Answer:
- D
Explanation:
increase while mechanical energy remains constant
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g
The time passed on earth is mathematically given as
t' = 24.79 hrs
<h3>What is the time passed on earth?</h3>
Generally, the equation for is time mathematically given as
![t' = \gamma t](https://tex.z-dn.net/?f=t%27%20%3D%20%5Cgamma%20t)
Where
![\gamma = Lorentz\ factor \\\\\gamma = 1/ \sqrt {(1 - v^2/c^2)}](https://tex.z-dn.net/?f=%5Cgamma%20%3D%20Lorentz%5C%20factor%20%5C%5C%5C%5C%5Cgamma%20%3D%201%2F%20%5Csqrt%20%7B%281%20-%20v%5E2%2Fc%5E2%29%7D)
![t' = t/ \sqrt {(1 - v^2/c^2)}](https://tex.z-dn.net/?f=t%27%20%3D%20t%2F%20%5Csqrt%20%7B%281%20-%20v%5E2%2Fc%5E2%29%7D)
Therefore
![t' = 24/ \sqrt {(1 - (0.25c/c)^2) }](https://tex.z-dn.net/?f=t%27%20%3D%2024%2F%20%5Csqrt%20%7B%281%20-%20%280.25c%2Fc%29%5E2%29%20%7D)
![t'= 24/ \sqrt {(1 - 0.25^2)](https://tex.z-dn.net/?f=t%27%3D%2024%2F%20%5Csqrt%20%7B%281%20-%200.25%5E2%29)
t' = 24.79 hrs
In conclusion, the time passed on earth
t' = 24.79 hrs
Read more about the time
brainly.com/question/28050940
#SPJ1
I did 25m/ 2.5s which is 10 m/s. not exactly sure tho