Q:

a water 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+3a. 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?

Accepted Solution

A:
h(t) = -46t² + 40t + 3We can think this graph by t being the x-axis and h being y-axisSo 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/4aLet's calculate:Δ = b² - 4.a.cΔ = 40² - 4.(-46).3Δ = 2152Yvertex = -2152/4.(-46)Yvertex = 2152/184Yvertex = 269/23Now we have the value of y we need to equal it to the equation269/23 = -46t² + 40t + 3-46t² + 40t + 3 - 269/23 = 0-46t² + 40t - 200/23 = 0Δ = 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 secondsh(2) = -46.2² + 40.2 + 3h(2) = -46.4 + 80 + 3h(2) = -184 + 83h(2) = -101Not sure how it's possible but it would be -101.