Answer:
sin 2θ = 1 θ=45
Explanation:
They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation
R = Vo² sin 2θ / g
Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.
We calculate the distance traveled for different angle
R = vo² Sin (2 15) /9.8
R = Vo² 0.051 m
In the table are all values in two ways
Angle (θ) distance R (x)
0 0 0
15 0.051 Vo² 0.5 Vo²/g
30 0.088 vo² 0.866 Vo²/g
45 0.102 Vo² 1 Vo²/g
60 0.088 Vo² 0.866 Vo²/g
75 0.051 vo² 0.5 Vo²/g
90 0 0
See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL
Answer:
Part a)

Part b)

Explanation:
Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

here we also use angular momentum conservation
so we have

also we know that the collision is elastic collision so we have

so we have

also we know

also we know

so we have


now we have


Part b)
Now we know that speed of the ball after collision is given as

so it is given as

Answer:
a.
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N


By Newton's second law



b.By Newton's second law

Substitute the values

Hence, the force in the coupling between the dogs and the sled=1058 N
Th answer is A sorry if this isn’t what your looking for
Answer:
wavelength.
Explanation:
One complete expression of a waveform beginning at a certain point, progressing through the zero line to the wave’s highest (crest) and lowest (trough) points, and returning to the same value as the starting point is called a is called wavelength. Its can be also defined as the distance between two successive crests or trough points in wave form.