h(t) = -46t² + 40t + 3
We can think this graph by t being the x-axis and h being y-axis
So we want the maximum value to y.
We know by math that the vertex of a parabola is (-b/2a, -Δ/4a)
So the y value of the vertex is -Δ/4a
Let's calculate:
Δ = b² - 4.a.c
Δ = 40² - 4.(-46).3
Δ = 2152
Yvertex = -2152/4.(-46)
Yvertex = 2152/184
Yvertex = 269/23
Now we have the value of y we need to equal it to the equation
269/23 = -46t² + 40t + 3
-46t² + 40t + 3 - 269/23 = 0
-46t² + 40t - 200/23 = 0
Δ = b² - 4.a.c
Δ = 40² - 4 . -46 . (-200/23)
Δ = 1600 - 4. -46 . (-200/23)
Δ = 0
There's 1 real root.
In this case, x' = x'':
x = (-b +- √Δ)/2a
x' = (-40 + √0)/2.-46
x'' = (-40 - √0)/2.-46
x' = -40 / -92
x'' = -40 / -92
x' = 0,43478260869565216
x'' = 0,43478260869565216
So, after approximately 0,4348 seconds the balloon will reach the highest point.
B) height after 2 seconds
h(2) = -46.2² + 40.2 + 3
h(2) = -46.4 + 80 + 3
h(2) = -184 + 83
h(2) = -101
Not sure how it's possible but it would be -101.
