A golf ball is hit and follows a parabolic path. It lands 220 yards from where it was hit and reaches a maximum height of 32 yards

Question

I need to know what the equation for this problem is and how high the ball was after it travels 100 yards 

Answer 1

I will set this up for you and you can work out the math.  When launched from the ground, and ignoring air resistance, the ball follows a path described by

 

(i)    y(x) = -[g/(2v₀²cos²θ)]x² + (tanθ)x 

 

y = height

x = horizontal distance

g = acceleration of gravity = 32 ft/s²

v₀ = initial launch velocity

θ = launch angle above the horizontal

 

The horizontal range of the ball is given by

 

(ii)    R = [v₀²sin(2θ)]/g

 

The maximum height is given by

 

(iii)    H = v₀²sin²θ/(2g)

 

You are given R = 220 yards = 660 ft, and H = 32 yards = 96 feet.

 

Given the values of R and H, and knowing g, you can find v₀ and θ from equations (ii) and (iii) (a system of two equations in two unknowns), and then substitute for v₀ and θ in equation (i).  Then just substitute x = 300 feet (i.e. 100 yards) into equation (i).  That will give the height y at a horizontal distance of x = 300 feet = 100 yards.

 

You should be able to finish the solution from here.  Feel free to post a comment if you have questions.

Answer 2

What do we know (or should we know) about parabolas?

 

They are represented by quadratic equations, or in other words, polynomials of degree 2.  So we are looking for the coefficients of a second degree polynomial.

We can assume the ground to be flat in this problem.  Let us graph the parabolic trajectory of the ball as seen from the side. We can consider the flat ground to correlate to the x-axis, the origin to be where the golfer is standing, and let the landing point be where x = 220. The parabola's value at any value of x between zero and 220 will represent the height of the ball above the ground as it passes a point x yards from the golfer.

 

The factored form of a parabola is y = A * (x-r₁) * (x-r₂) .  We know that this parabola is upside down, so A must be negative.  We also the maximum height of an upside-down parabola occurs at its axis of symmetry, which is always halfway between the roots.  Roots are places where the equation equals zero, and in this case those places where the height of the ball above the ground equals zero are the tee and the landing point.  After the ball hits the ground it might bounce or roll or squish into the mud ... in any case it will no longer be following the parabolic path, but whatever happens to the ball upon impact is not our concern in this problem.We just want to know what equation describes the parabolic portion of the ball's path.

 

Root 1 must be x=0, root 2 must be x=220.  We also know that the axis of symmetry must occur at x=110 yards.

Therefore, we set y = 32 when x = 110 so that we can solve for A.

y = A * (x-0)*(x-220)

Substituting in y=32 yards, x = 110 yards:

32 = A*(110)*(110-220)

Simplifying:

32 = A*(110)*(-110)

Cancelling factors of two:

A = -32/(121*100) = -16/(121*50) = -8/(121*25) = -8/((11^2)*(5^2)) = -8/(55^2)

Simplifying the denominator:

55^2 = (50+5)^2 = 50^2 + 2*50*5 + 5^2 = 2500 + 500 + 25 = 3025

 

therefore A = -8/3025

 

y = (-8/3025)*x^2 - ((220*8)/3025)*x

y = (-8/3025)*x^2 - ((11*4*5*8)/(5*5*11*11))*x

y = (-8/3025)*x^2 - ((4*8)/(5*11))*x

y = (-8/3025)*x^2 - (32/55)*x

 

Optional. Put coefficients into decimal approximation form

 

Complete the problem by solving for y when x = 100.  

 

Bonus question: Is there another point where the height of the ball will be the same as your answer and if so, where?

 

 

 

Answer 3

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