Person A , Person B.

They stood 3 m apart and they were also 1.5 m vertically from the ground when Person A decided to throw a ball to his friend Person B. The path of the ball formed a parabola of the form 𝑦 = π‘Žπ‘₯^2 + 𝑏π‘₯ + 𝑐, where x represents the horizontal distance the ball travelled and y represents the height above the ground. determine the function when π‘₯ = 1 π‘‘β„Žπ‘’ π‘ π‘™π‘œπ‘π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘Žπ‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘™π‘™ π‘€π‘Žπ‘  1.

Using this information and using calculus methods only, determine the maximum height of the ball.

1 answer

since you have y = ax^2 + bx +c
y' = 2ax + b
so
2a+b = 1
b = 1-2a
So now you have
y' =ax^2 + (1-2a)x
max height is reached when x = (2a-1)/(2a) = 1 - 1/(2a)
y = ax^2 + (1-2a)x + 3/2
At that point, y = -(4a^2-10a+1)/a^2

If you have to come up with actual numeric values for a,b,c, then the question now is: what is a?
Recall that when the ball was launched, at an angle ΞΈ,
x = v cosΞΈ t
y = v sinΞΈ t - 4.9t^2 + 1.5
That means that at time t,
tanΞΈ = (9.8t - v cosΞΈ)/(v sinΞΈ)
See what you can do with that to find the angle ΞΈ and initial velocity v.
Then the equation is
y = -g/(2(v cosΞΈ)^2) x^2 + tanΞΈ x + 1.5