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# 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

- 9 years agoFavorite 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 .

- marnieLv 45 years ago
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)

- ?Lv 79 years ago
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

- Anonymous9 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

- 9 years ago
29.5 meters. Replace t by 1. Substitute and you get the naswer. Ok?

Source(s): Prior knowledge.