Solution:
We can use the equations of motion to derive these
equations.
First, we know that the force acting on the object is given
by:
$F = ma$
Where F is the force in Newtons, m is the mass in kg, and a
is the acceleration in m/s^2.
In this case, the force acting on the object is P kg
weights, so we can substitute that in:
$P = ma$
Solving for a, we get:
$a = \frac{P}{m}$
Next, we know that the distance x traveled by the object is
given by:
$x = \frac{1}{2}at^2$
Substituting in our expression for a, we get:
$x = \frac{1}{2} \left(\frac{P}{m}\right)t^2$
Simplifying, we get:
$x = \frac{Pt^2}{2m}$
This is our first equation.
Next, we know that the final velocity of the object is given
by:
$V = u + at$
Where V is the final velocity in m/s, u is the initial
velocity (which is 0 in this case), and t is the time in seconds.
Substituting in our expression for a, we get:
$V = \frac{Pt}{m}$
Solving for t, we get:
$t = \frac{Vm}{P}$
Substituting this into our expression for x, we get:
$x = \frac{P}{2m} \left(\frac{Vm}{P}\right)^2$
Simplifying, we get:
$x = \frac{V^2m}{2P}$
This is our second equation.
Finally, we know that the acceleration due to gravity is g =
9.81 m/s^2. We can substitute this in to get our final expressions:
$x = \frac{gt^2P}{2m} = \frac{g}{2m}
\left(\frac{Vm}{P}\right)^2 = \frac{gV^2m}{2P^2}$
Simplifying, we get:
$x = \frac{gt^2P}{2m} = \frac{V^2m}{2gP}$
These are the equations we wanted to show.