Why is 45 degrees the best angle of trajectory?

Question

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Answer 1

Base on equation of displacement of trajectory, 45 degree is the best angle to launch an object to the farthest distant.

Answer 2

It's only the best trajectory if you have a launcher that has a fixed launch velocity regardless of launch angle and your projectile has no lift and you are interested in getting the maximum horizontal displacement on a level plane.

Answer 3

By "best angle" you clearly mean the angle at which a projectile will travel farthest, all other factors being held constant.

To understand this, first assume that a projectile is fired at an angle theta to the horizontal at initial velocity v0. Then, its initial vertical velocity is v0*sin(theta) and its initial horizontal velocity is v0*cos(theta). But it is a known fact that a projectile fired vertically at velocity v will hit the ground after a period of time given by 2v/g. So we can conclude that the projectile above will hit the ground at time 2v0*sin(theta)/g.
Now, if we neglect air resistance (as you are supposed to do), there is no force to retard the horizontal velocity v0*cos(theta) until it hits the ground. Therefore the distance travelled will be (v0*cos(theta)) * (2v0*sin(theta)/g). If we multiply these, we get 2*v0^2*cos(theta)*sin(theta)/g. Trigonometry indicates that 2*cos(theta)*sin(theta) = sin(2*theta), so this simplifies to the following:

A projectile fired at an initial velocity of v0 at an angle theta will travel a horizontal distance of v0^2 * sin(2*theta) / g.

We want to maximize the value of this distance. Since we are keeping the initial velocity constant and "g" simply is a constant, we want to maximize sin(2*theta). The maximum value of the sine function is 1, for 90 degrees. Therefore sin(2*theta) is maximized for 45 degrees.

Answer 4

When you write the equations of motion for a projectile in a uni-directional gravitational field you get

R = Vo²sin2Θ/g

Setting the derivative of this function = 0 gives

sin2Θ = 1 → Θ = 45°

Answer 5

It's based on the formula which gives range as a function of launch angle:

R = V²sin(2θ)/g

To maximize the range, you need to maximize the value of sin(2θ). That occurs when θ = 45 degrees.

Answer 6

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Answer 7

how far something goes depends on its horizontal speed and how long it travels. d=v * t
You maximize the product of the speed and the time.
That is to say how fast it is going and how long it stays in the air.