A rocket is launched in the air. Its height in feet is given by h, equals, minus, 16, t, squared, plus, 128, th=−16t 
2
 +128t where tt represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point?

To find the time at which the rocket reaches its highest point, we need to find the vertex of the quadratic equation:

\[
h(t) = -16t^2 + 128t
\]

This equation is in the standard form of a quadratic: \( h(t) = at^2 + bt + c \), where \( a = -16 \) and \( b = 128 \). The time at which the maximum height occurs for a quadratic \( ax^2 + bx + c \) is given by:

\[
t = -\frac{b}{2a}
\]

Let's substitute the values of \( a \) and \( b \):

\[
t = -\frac{128}{2 \times (-16)} = \frac{128}{32} = 4
\]

Thus, the rocket reaches its highest point **4 seconds** after launch.

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. What is the maximum height the rocket reaches?
2. How long is the rocket in the air before it hits the ground?
3. What is the height of the rocket after 2 seconds?
4. What is the equation of the rocket's path if the launch velocity changes?
5. How would air resistance affect the rocket's maximum height?

**Tip:** In any quadratic equation describing motion, the vertex gives the maximum (or minimum) point, which is useful in physics problems.

A rocket is launched in the air. Its height in feet is given by h = -16t^2 + 128t where t represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point?

Discover how to find the time at which a rocket reaches its highest point using the quadratic equation h = -16t^2 + 128t. This detailed solution covers key algebra concepts, the vertex formula, and solving for maximum height. Ideal for students in Grades 9-11.

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