What is the optimum angle to throw a snowball for maximum horizontal displacement? (Ignore air resistance, assume the snowball is thrown level with the ground. The angle is measured from the ground up)

Shortest answer:

45 degrees

 

Short Answer:

If the snowball is thrown at velocity v, at angle θ

The vertical component is vsin(θ)

The horizontal component is vcos(θ)

v = u + at            t = v/a     (u = 0)

Airtime = 2vsin(θ)/g

(horizontal distance = x)

x = vcos(θ) * 2vsin(θ)/g        =     2v^2sin(θ)cos(θ)/g

2sin(θ)cos(θ) = sin(θ)

X= 20.5v^2sin(2θ)/g         =  v^2sin(2θ)

Dx/dθ = 2v^2cos(2θ)

2v^2cos(2θ) = 0

2θ = cos^-1(0)     (arc cosine)

 θ = 0.5cos^-1(0)

θ = 45 degrees

Yay

 

 

 

Long, but fully explained answer:

The first thing you want to do with ANY mechanics problem is draw a diagram. Unfortunately, I cant on this, so try and draw one yourself.

Our diagram should show the velocity vector (arrow) that we are throwing the snowball with, and the angle between that arrow and the ground (θ).

From our diagram, we can use some trigonometry to see that if the velocity we throw the ball at is ‘v’,

The vertical component is vsin(θ)

The horizontal component is vcos(θ)

For projectile questions about distance travelled, we use the vertical component to find the time the object is in the air, and multiply this by the horizontal velocity, as distance = speed * time

In this problem, we want to know something more general, the angle of projection which will result in the greatest horizontal displacement. If we throw the ball completely horizontal, the vertical component will equal zero and therefore the ball will (theoretically) not travel at all, as it has zero airtime.

Throwing the snowball with a large angle will give it lots of airtime, and therefore time to travel, while throwing with a smaller angle will give it a larger horizontal velocity, meaning it will cover more horizontal distance in a shorter time. We need a compromise of the two. But for this problem, you don’t even need to think about that if you’re on top of your maths.

Let’s think of this mathematically:

First let’s consider the snowball’s airtime:

Its airtime is equal to the time it takes to reach its maximum height and then fall back down again. If the snowball is launched from the ground, these two are equal.

Our best friend, SUVAT tells us that:

v = u + at

u (the initial velocity) is zero, and so we can rearrange the equation to:

t = v/a

We can use SUVAT because there is no air resistance and so the acceleration is constant. We can only ever use SUVAT if the acceleration is constant – for every projectile question you will face, it will be)

Remember that in this case, we are dealing with the vertical velocity, and that the airtime is 2* the time to the maximum height. Then we have:

Airtime = 2t   =   2vsin(θ)/g 

(SUVAT’s v means the velocity in the direction its calculating, so in our case it equals vsin(θ), the acceleration is the acceleration due to gravity, g)

We now have both the horizontal velocity (vcos(θ)) and  the time (airtime) that we need to find the distance. Let us call the distance ‘x’:

X = vcos(θ) * 2vsin(θ)/g        =     2v^2sin(θ)cos(θ)/g

(v^2 = v squared) (distance = speed * time)

Great! Now we have a general formula for the distance travelled by ANY projectile (although this equation is only valid as long as there is no air resistance or external forces, such as wind, and the equation does not account for any bouncing)

Imagine throwing a snowball at various angles. If we throw the snowball completely horizontal, the vertical component will equal zero and therefore the ball will (theoretically) not travel at all (if launched from ground height), while if we throw the snowball directly up, it will go up, come back down and hit one of us in the face. Somewhere in between these two angles (0 and 90 degrees) we will find our optimum angle.

So we really want to know how the maximum distance travelled by the snowball changes as we change the angle, and to do this, we must differentiate our general distance equation with respect to its angle. If we were lazy, we could use trial and error, but that’s no fun.

We can sensibly assume that v and g are constants (the gravitational field strength will barely change with this relatively small altitude change, and there is no air resistance, meaning that the horizontal component of the velocity, v, is constant)

But we have a problem. I don’t know how to differentiate sin(θ)cos(θ). Perhaps we can change what we are differentiating. Double angle formulae tells us that 2sin(θ)cos(θ) = sin(2θ), meaning that sin(θ)cos(θ) = 0.5sin(2θ). Let’s substitute this in:

x= 20.5v^2sin(2θ)/g         =  v^2sin(2θ)

Now we can differentiate it:

(differential of sin(aθ) = acos(aθ) and remember that v^2 is a constant)

dx/dθ = 2v^2cos(2θ)

Fantastic! Now we have an equation which tells us how the maximum distance travelled by the snowball varies with the angle.

We want to know when the distance travelled is at a maximum, so we will use stationary points. Stationary points tell us the maximum or minimum of a function. We have a stationary point when the derivative = 0.

Then we get:

2v^2cos(2θ) = 0

Lets divide everything by 2v^2:

cos(2θ) = 0 

and therefore:

2θ = cos^-1(0)     (arc cosine)

 θ = 0.5cos^-1(0)

θ = 45 degrees!

Yay!

Because we used stationary points, this could be a minimum. We could check this, but logic tells us that a snowball launched at 45 degrees is a sensible answer. If you do want to go further though, because cosine is periodic, the other solutions are θ = 45,135,225,315,405 and so on. 135 degrees is essentiallt the same angle, but behind us, 225 and 315 degrees are straight into the ground and 405 degrees means that we did a backflip and then aimed at 45 degrees, as 405-360 = 45 (360 degrees is a full circle).

So we know that our answer is correct. Note that there are an infinite amount of solutions to θ = 0.5cos^-1(0) because cosine goes on forever. (enter y=cos(2x) on a graphing website if you’re struggling to visualize this, and look for when it intercepts the x axis – when it is equal to zero)

All of these extra solutions are like us doing more and more backflips before we throw our snowball. (each ‘backflip’ = 360 degrees)