4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector

where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with


The Coyote hits the ground when
:

4b. Here we evaluate
at the time found in (4a).

5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is

We find the shell hits the ground at

5b. The horizontal component of the bullet's position vector is

where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:

A because the girl in that instant is not moving up or down so
( up forces)=(down forces )
The up forces is the tension of the rope and down forces us mg -the gravitational force on the girl by the earth
Answer:
Newtons.
Explanation:
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
Where;
F represents force measured in Newton.
m represents the mass of an object measured in kilograms.
a represents acceleration measured in meter per seconds square.
Newtons is a measurement we use to determine the amount of force used to move an object by a simple machine. It is the International System of Units (SI) used to measure force and has a symbol of N.
Basically, it was named after Sir Isaac Newton based on his fundamental works in the field of mechanics (motions).
mass of the bottle in each case is M = 0.250 kg
now as per given speeds we can use the formula of kinetic energy to find it
1) when speed is 2 m/s
kinetic energy is given as


2) when speed is 3 m/s
kinetic energy is given as


3) when speed is 4 m/s
kinetic energy is given as


4) when speed is 5 m/s
kinetic energy is given as


5) when speed is 6 m/s
kinetic energy is given as

