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 molecular geometry of both F2 and HF is linear.There are only two atoms which are covalently bonded and thus, the bonding scheme with the atoms looks like this;
F --- F
H---F
So, both are linear.
Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

here
speed of first plane is 700 mi/h at 31.3 degree


speed of second plane is 570 mi/h at 134 degree


now the relative velocity is given as


now the distance between them is given as



so the magnitude of the distance is given as

miles
so the distance between them is 2985.6 miles