An interesting projectile motion problem

Question

I found this problem on the internet:

Conditions are:

No air resistance and the ball is a point of mass.

Also, the size of the soccer goal is slightly different from the standard one.

The questions:

What is the distance "a"? What is the angle α? What was the initial speed of the ball for those two free kicks?

Task: A free kick is taken from point "K" at a certain distance "a" from a soccer goal. Point "K" is located on the perpendicular bisector of the goal line.

The ball is kicked at an angle α just towards the middle of the upper goal post, but after the movement in time t(1) it falls to the ground near the middle of the goal line.

The second time they kick the ball harder. The ball is kicked at an angle 2α, which after t(2) hits the middle of the upper goal post.

t(1)=3s , t(2)=3,5s

Thank you for any help!

Answer 1

The answer to this question really depends on whether we can assume the initial velocity of the two kicks are the same with only the angle being different. From the reading of the question is appears this is the case though it is a little ambiguous. For the case where the velocities are not equal this becomes a much more difficult problem.

The case where v₁ and v₂ are equal

kick 1

The horizontal displacement is given by x = v₁Cos(α)t₁

the vertical displacement is given by y = v₁Sin(α)t₁ - 1/2 * g * t₁²

Thus,

a = v₁Cos(α)t₁

At time t₁ the ball hits the ground so it's displacement is 0 or y = 0.

0 = v₁Sin(α)t₁ - 1/2 * g * t₁²

we can solve for v₁

v₁ = 1/2 * g/Sin(α)

Kick 2

a = v₂Cos(2α)t₂

at time t₂ the ball is a height h above the ground. (we define h as the height of the top goal post)

h = v₂Sin(2α)t₂ - 1/2 * g * t₂²

we are assuming that v_{1 =} v₂ thus,

h = 1/2 * g/Sin(α) * Sin(2α)t₂ - 1/2 * g * t₂²

h = 0.5*gt₂ * Sin(2α)/Sin(α) - 1/2 * g * t₂²

We can make use of the double angle identity for Sin(2α)/Sin(α) = 2Cos(α)

h = g*Cos(α) - 1/2 * gt₂²

Solve for α

α = cos^-1(h + 1/2 * gt₂²) SOLUTION for α

We can use the solution for v₁ in the equation for a (kick 1)

a = 1/2 * g * Cos(α)/Sin(α) * t₁

a = 1/2 * g * 1/tan(α) * t₁

the tan of an inverse cosine function, tan(cos^-1(x)) = sqrt(1-x²)/x

here x is equal to h + 1/2 * gt₂² from our solution for α

so 1/tan(α) = x/sqrt(1 - (h + 1/2 * gt₂²)²) and

a = 1/2 * g * x/sqrt(1 - (h + 1/2 * gt₂²)²) * t₁ SOLUTION FOR 'a'

v₁ = v₂ = 1/2 * g/Sin(α)

the sin of an inverse cosine, Sin(Cos^-1(x)) = sqrt(1-x²)

v₁ = v₂ = 1/2 * g/sqrt(1 - (h + 1/2 * gt₂²)²) SOLUTION FOR velocity

As we know t₁ and t₂, for any given h (height of the goal post) we can calculate the anlge α, the distance 'a' and the velocity v_1,2 of the kicks.