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# Math question can anyone please help?

A person standing close to the edge on top of a 144-foot building throws a baseball vertically upward. The quadratic function models the ball's height above the ground, s(t)= - 16t^2+64t+144, in feet, t seconds after it was thrown. After how many seconds does the ball reach its maximum height? Round to the nearest tenth of a second if necessary.

### 4 Answers

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• Quadratic Equation solution for f(x) or y = ax² +bx + c is: x = -b/2a +-√(b² -4ac)/2a   (Vertex ± offset)

So t = -b/2a or -64/-32

t = 2.0 sec

***

If you know calculus, you take 1st differential = 0

slope = 0 for the vertex

• When you have a quadratic, of the form ax² + bx + c, the equation for the line of symmetry is:

x = -b/(2a)

If you need help remembering this, it's the quadratic formula, but without the ±√ part.

In your case the variable is t (time), but its the same thing:

t = -b/(2a)

Plug in your values:

a = -16

b = 64

t = -(64) / (2 * -16)

t = -64/-32

t = 2

That is the vertex of your parabola and since the leading coefficient (a = -16) is negative, the parabola is downward facing. That means the vertex is a maximum.

Answer:

It reaches its maximum height after 2 seconds.

• s(t) = -16t² + 64t + 144

s(t) = -16(t² - 4t) + 144

s(t) = -16(t² - 4t + 4) + 144 + 16(4)

s(t) = -16(t - 2)² + 208

The maximum height is reached at t - 2 = 0.

t = 2

The baseball reaches a maximum height of 208 feet at two seconds.

• s(t) = –16t² + 64t + 144

diff and set equal to 0 to get max, ie, when v = 0

s' = v = 0 = –32t + 64

t = 2 s

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