Answer:
Velocity.
Explanation:
Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.
As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:
Horizontal range: As per expression:
R= (
*sin2θ)/g
the range depending on the square of the initial velocity.
Maximum height: As per expression:
H= ( * 
θ
)/2g
the maximum distance also depends upon square of the initial velocity.
Player 2 because moment is mass times acceleration and since they are all going the same speed. Speed doesn't matter so the only thing that is left is mass/ weight and he has the most
Answer: 
Explanation:
If we make an analysis of the net force 
of the rock that was thrown upwards, we will have the following:
(1)
Where:
is the force with which the rock was thrown
is the weight of the rock
Being the weight the relation between the mass 
of the rock and the acceleration due gravity 
:
(2)
(3)
Substituting (3) in (1):
(4)
(5) This is the net Force on the rock
On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:
(6)
Finding the acceleration 
:
(7)
(8)
Finally:
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 = ![\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bg%7D%20%5C%20%5B%28v_%7Bo%7D%20-%20v%29%20%20-%20%28%20-%20v_%7Bo%7D%20%20-%20v%29%20%5D)
Δt = 2 v₀ / g
