Awater balloon is tossed into the air with an upward velocity of 40 ft/sits height h(t) in ft after t seconds is given by the function h(t) =-46t^2+40t+3

a. after how many seconds will the balloon be at the highest point above the ground?

b. what will the height be at t = 2 seconds?

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.

55/i dont know i do it for the points

step-by-step explanation: