Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

# Quadratic Function Question NEED HELP ASAP!?

A ball is thrown into the air. Its height,h, in meters, after t seconds is h=-4.9(t-1.8)^2+17. What is the height of the ball when it was released? Correct your answer to one decimal place.

### 5 Answers

Relevance
• Favorite Answer

When it was released t=0. That makes your expression -4.9(-1.8)^2 + 17. A calculator will give you the answer: 1.1 meters .

• The vertex type of a parabola is y = a(x - c)^2 + d, where (c , d) is the vertex a factor on the parabola: 1) (-3 , 1) = (x , y) ==> 1 = a(-three - c)^2 + d (1 - d) = a(-(3 + c))^2 = a(3 + c)^2 a = (1 - d) / (3 + c)^2 so now the equation of the parabola is y = [(1 - d) / (3 + c)^2] (x - c)^2 + d let's plug in x = 0 to find the y-intercept: y = [(1 - d) / (3 + c)^2]c^2 + d so in phrases of (c , d), the point (0 , [(1 - d) / (3 + c)^2]c^2 + d) is the y-intercept likewise for #2, except replace (-three , 1) with (a , b) b = A(a - c)^2 + d ==> A = (b - d) / (a - c)^2 y = [(b - d) / (a - c)^2](x - c)^2 + d the y-intercept can be (zero , [(b - d) / (a - c)^2]c^2 + d)

• the height when the ball is released is when t = 0

h = - 4.9 (-1.8)^2 + 17

h = - 4.9 (3.24) + 17

h = - 15.876 + 17

h =1.124 m = 1.1 m

• Anonymous
9 years ago

ball realesed at 0 seconds: so t=0

just put t=0 in the equation and resolve it

h=-4.9(-1.8)^2+17 = 1.1meters

• 29.5 meters. Replace t by 1. Substitute and you get the naswer. Ok?

Source(s): Prior knowledge.
Still have questions? Get your answers by asking now.