Question
A rocket is launched in the air. Its height in feet is given by h, equals, minus, 16, t, squared, plus, 112, th=−16t 2 +112t where tt represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point?
Answers
GPT-4o mini
The height of the rocket as a function of time is given by the equation:
\[
h = -16t^2 + 112t
\]
To find the time at which the rocket reaches its highest point, we can use the formula for the vertex of a quadratic function, since the equation is in the standard form \(h(t) = at^2 + bt + c\). The time at which the maximum height occurs can be found using the formula:
\[
t = -\frac{b}{2a}
\]
In this case, \(a = -16\) and \(b = 112\).
Plugging in the values:
\[
t = -\frac{112}{2 \times -16} = -\frac{112}{-32} = \frac{112}{32} = 3.5
\]
Thus, the rocket reaches its highest point at \(t = 3.5\) seconds.
\[
h = -16t^2 + 112t
\]
To find the time at which the rocket reaches its highest point, we can use the formula for the vertex of a quadratic function, since the equation is in the standard form \(h(t) = at^2 + bt + c\). The time at which the maximum height occurs can be found using the formula:
\[
t = -\frac{b}{2a}
\]
In this case, \(a = -16\) and \(b = 112\).
Plugging in the values:
\[
t = -\frac{112}{2 \times -16} = -\frac{112}{-32} = \frac{112}{32} = 3.5
\]
Thus, the rocket reaches its highest point at \(t = 3.5\) seconds.