Question

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 = 𝑎𝑥^2 + 𝑏𝑥 + 𝑐, where x represents the horizontal distance the ball travelled and y represents the height above the ground. determine the function when 𝑥 = 1 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙 𝑤𝑎𝑠 1.

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

Answer 1

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 θ,
x = v cosθ t
y = v sinθ t - 4.9t^2 + 1.5
That means that at time t,
tanθ = (9.8t - v cosθ)/(v sinθ)
See what you can do with that to find the angle θ and initial velocity v.
Then the equation is
y = -g/(2(v cosθ)^2) x^2 + tanθ x + 1.5