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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
- JimLv 72 months ago
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
- PuzzlingLv 72 months ago
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.
- PopeLv 72 months ago
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.
- billrussell42Lv 72 months ago
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
