Question
An object is thrown into the air. Graph the situation using the formula y=−4t2+10t . What is the maximum height the object will reach in feet?
Answers
GPT-4o mini
To find the maximum height of the object described by the equation \( y = -4t^2 + 10t \), we need to identify the vertex of the parabola represented by this quadratic equation. The formula for a parabola in the standard form \( y = at^2 + bt + c \) gives us the coefficients:
- \( a = -4 \)
- \( b = 10 \)
- \( c = 0 \) (though \( c \) does not affect the maximum height we are looking for)
The maximum height occurs at the vertex, and the time \( t \) at which the vertex occurs can be found using the formula:
\[
t = -\frac{b}{2a}
\]
Substituting in our values for \( a \) and \( b \):
\[
t = -\frac{10}{2 \times -4} = -\frac{10}{-8} = \frac{10}{8} = \frac{5}{4} = 1.25 \text{ seconds}
\]
Now we substitute \( t = 1.25 \) back into the original equation to find the maximum height \( y \):
\[
y = -4(1.25)^2 + 10(1.25)
\]
Calculating \( (1.25)^2 \):
\[
(1.25)^2 = 1.5625
\]
Now substituting back into the equation:
\[
y = -4(1.5625) + 12.5
\]
\[
y = -6.25 + 12.5
\]
\[
y = 6.25 \text{ feet}
\]
Thus, the maximum height the object will reach is **6.25 feet**.
- \( a = -4 \)
- \( b = 10 \)
- \( c = 0 \) (though \( c \) does not affect the maximum height we are looking for)
The maximum height occurs at the vertex, and the time \( t \) at which the vertex occurs can be found using the formula:
\[
t = -\frac{b}{2a}
\]
Substituting in our values for \( a \) and \( b \):
\[
t = -\frac{10}{2 \times -4} = -\frac{10}{-8} = \frac{10}{8} = \frac{5}{4} = 1.25 \text{ seconds}
\]
Now we substitute \( t = 1.25 \) back into the original equation to find the maximum height \( y \):
\[
y = -4(1.25)^2 + 10(1.25)
\]
Calculating \( (1.25)^2 \):
\[
(1.25)^2 = 1.5625
\]
Now substituting back into the equation:
\[
y = -4(1.5625) + 12.5
\]
\[
y = -6.25 + 12.5
\]
\[
y = 6.25 \text{ feet}
\]
Thus, the maximum height the object will reach is **6.25 feet**.