Physics Assignment: Solutions for Projectile Motion, Growth, and More

This physics assignment solution provides detailed explanations and solutions to various physics problems. The assignment covers topics such as projectile motion, where the trajectory of objects is analyzed using equations of motion and velocity components. It also includes the application of the continuity equation to determine the flow of fluids and explores the concept of bacterial growth, utilizing coefficients derived from the least squares method. The solutions involve breaking down complex scenarios into manageable steps, applying relevant formulas, and deriving equations to solve for unknown variables. The assignment emphasizes the use of calculus for optimization and the understanding of physical principles to address real-world scenarios, making it a comprehensive resource for physics students seeking to understand and solve a range of problems.

1)
To get the best performance we first need to take the log of the given equation,
P=0.1298 a2+ b
√4.123 cd
log P=2 loga+log 0.1298+ 1
b ¿
Now this can be solved using partial differentiation and equating each of the four equations we
obtain,
∂ log P
∂ a =0
∂ log P
∂b =0
∂ log P
∂ c =0
∂ log P
∂ d =0
2)
We can use the concept of projectile motion to solve it. Since the trajectory is a projectile and the
initial velocity and the angle of projection are given, we can use the normal equations of projectile
motion to solve the problem
The velocity will be broken down in to two components one horizontal and the other vertical. The
horizontal component is 25cos16 and the vertical is 25sin16. Then we should find out the time for
which
y=25sin 16 ° t−1
2 g t2
Is satisfied and from the t which we will find out
x=25 cos 16 ° t
3)
The techniques of equation of continuity should be used to solve the problem. A continuity equation
in physics is an equation that describes the transport of some quantity. So the amount of water
flowing per min in to the cylinder if is greater than the amount of water per flowing out per min is
through the hole at the bottom of the cylinder then it will over flow otherwise if it less then it will
become empty
4)

We can use the concept of projectile motion to solve it. Since the trajectory is a projectile and the
initial velocity and the angle of projection are given, we can use the normal equations of projectile
motion to solve the problem
For the first one,
The velocity will be broken down in to two components one horizontal and the other vertical. The
horizontal component is 60cos25 and the vertical is 60sin25. Then we should find out the time for
which
y=60sin 25 ° t −1
2 g t2
0=60 sin 25° t− 1
2 g t2
We will get two values of time one at 0 and another at
t1=0 , t2 =120 sin 25 °
g
So at midway then
tmidway= 60 sin 25 °
g
For the second one,
For the first one,
The velocity will be broken down in to two components one horizontal and the other vertical. The
horizontal component is 60cos35 and the vertical is 60sin35. Then we should find out the time for
which
y=60sin 3 5 ° t− 1
2 g t 2
0=60 sin 35 ° t −1
2 g t2
We will get two values of time one at 0 and another at
t1=0 , t2 =120 sin 3 5°
g
So at midway then
tmidway= 60 sin 3 5°
g
We need to find out tmidway for both of them and again find out the value of y for that time and see if
the y value are same.

5)
It will be set when the first order differentiation of g will be zero
dg
dc =0
This will provide the equation of c from which we can calculate the value of c
6)
bacterial growth=a ×temperature +b × humidity+c × time+ d
Now this coefficients of a, b, c and d are found out using the least squares method used in the
statistics.