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
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 π¦ = ππ₯^2 + ππ₯ + π, where x represents the horizontal distance the ball travelled and y represents the height above the ground. determine the function when π₯ = 1 π‘βπ π ππππ ππ π‘βπ π‘ππππππ‘ π‘π π‘βπ πππ‘β ππ π‘βπ ππππ π€ππ 1.
Using this information and using calculus methods only, determine the maximum height of the ball.
1 answer