If we drew this out and set the tee box at (0, 0), we can solve for c:

y = ax^2 + bx + c

0 = a(0)^2 + b(0) + c

c = 0

So we’re left with y = ax^2 + bx. We still don’t know what a and b are, but we can find that too. The derivative of this equation gives us the slope of the tangent at any point on the curve. So taking the derivative, we have:

y = ax^2 + bx

y’ = 2ax + b

Now for some trig – we have an angle of 25 degrees at (0, 0). The slope at that point is tan 25 (tangent is y/x, which is our slope for this curve), or 0.466. We can plug in this to find b:

y’ = 2ax + b

0.466 = 2a(0) + b

b = 0.466

We can find a now if we go back to our original equation, using (301.5, 0) as our other point on the parabola:

y = ax^2 + bx

0 = a(301.5)^2 + 0.466(301.5)

0 = 90902.25a + 140.499

-140.499 = 90902.25a

a = -0.00155

The negative makes sense because the parabola is inverted. So our final equation becomes:

y = -0.00155x^2 + 0.466x

Again, assuming that nothing affects ball flight, it would makes sense that at half of 301.5, or 150.75 meters, the ball would peak. Given that, we can plug in and solve:

y = -0.00155x^2 + 0.466x

y = -0.00155(150.75)^2 + 0.466(150.75)

y = -34.088 + 70.2495

y = 36.1615 m

Alternatively, you can take the derivative of the equation we found and solve for a 0 slope to find at what x coordinate the ball peaks:

y = -0.00155x^2 + 0.466x

y’ = -0.0031x + 0.466

0 = -0.0031x + 0.466

0.0031x = 0.466

x = 150.322 m

Because of rounding in finding a and b, this is slightly off of our 150.75 m deduction, but from here the process is the same – plug in the 150.322 m into the other equation to figure out how high the ball reaches.

The answer makes sense when you think about it – 36.1615 m is about 119 feet – that’s about the right height if you ever watch golfers teeing off.

u m/sec be the initial velocity of the ball,

t sec be the flight time,

h metres be the height at time t.

Resolving horizontally and vertically:

301.5 = ut cos(25) …(1)

h = ut sin(25) – gt^2 / 2 …(2)

Putting h = 0 in (2):

0 = ut sin(25) – gt^2 / 2

t(u sin(25) – gt/2) = 0

Therefore:

t = 0 (initially)

and

t = 2u sin(25) / g …(3)

when the ball reaches the ground again.

Substituting for t from (3) in (1):

301.5 = 2u^2 sin(25)cos(25) / g

u^2 = 301.5g / 2sin(25)cos(25) …(4)

If H is the maximum height reached, then:

0 = u^2 sin^2(25) – 2gH

H = u^2 sin^2(25) / 2g …(5)

Substituting for u^2 from (4) in (5):

H = 301.5 tan(25) / 4

H = 35.15m.