SOLUTION: If a ball is thrown upward from the top of a building 30m high with an initial velocity of 10m/s, then the height (h) above ground t seconds later will be h = 30 + 10t

SOLUTION: If a ball is thrown upward from the top of a building 30m high with an initial velocity of 10m/s, then the height (h) above ground t seconds later will be h = 30 + 10t - 5t^2

Question 1203081: If a ball is thrown upward from the top of a building 30m high with an initial velocity of 10m/s, then the height (h) above ground t seconds later will be
h = 30 + 10t - 5t^2
During what time interval will the ball be at least 15m above the ground?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
set the equation equal to 15 to get:
30 + 10t - 5t^2 = 15
subtract 15 from both sides of the equation to get:
15 + 10t - 5t^2 = 0
rearrange the equation in descending order of degree to get:
-5t^2 + 10t + 15 = 0
multiply both sides of the equation by -1 to get:
5t^2 - 10t - 15 = 0
divide both sides of the equation by 5 to get:
t^2 - 2t - 3 = 0
factor the equation to get:
(t-3) * (t+1) = 0
solve for t to get:
t = 3 or t = -1
t has to be positive so t = 3.
when t = 3, -5t^2 + 10t + 15 becomes -45 + 30 + 15 whih becomes 0, as it should.
go back to the original equqtion of h = 30 + 10t - 5t^2 and replace t with 3 to get:
h = 30 + 30 - 45 = 15.
that's the height when t = 3.
that's your solution.
here's a graph of the result.

in the graph, y takes the place of h and x takes the place of t.

the value of h will be at least 15 feet above the ground from t = 0 to t = 3.
in the graph, that's from x = 0 to x = 3.

while the graph does show negaive values of t, that is not part of the domain since t has to be greater than or equal to 0 (time is not negative).

to see that clearer, just draw a horizontal line at y = 15 on the graph.

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