SOLUTION: The height h metres of a ball at time t second after it is thrown up into the air Is given by the expression h= 1+ 15t - 5t^2 (i) Find the times at which the height is 11m (ii)

Question 1200805: The height h metres of a ball at time t second after it is thrown up into the air Is given by the expression
h= 1+ 15t - 5t^2
(i) Find the times at which the height is 11m
(ii) Use your calculator to find the time when the ball hits the ground
(iii) what is the greatest height the ball reaches
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The height h metres of a ball at time t second after it is thrown up into the air Is given by the expression
h= 1+ 15t - 5t^2
(i) Find the times at which the height is 11m
h= 1+ 15t - 5t^2 = 11
5t^2 - 15t + 10 = 0
t^2 - 3t + 2 = 0
(t-1)*(t-2) = 0
t = 1, t = 2 seconds.
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(ii) Use your calculator to find the time when the ball hits the ground
h= 1+ 15t - 5t^2 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=245 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.0652475842498529, 3.06524758424985. Here's your graph:

Ignore the negative value.
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(iii) what is the greatest height the ball reaches
The vertex of the parabola is at t = -b/2a
t = -15/-10 = 1.5 seconds
h(1.5) = 12.25 meters
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