A ball is thrown so that its height above the ground seconds after it has been thrown is given by the function (h)t=20t-4.9^2-2. After 3 seconds, the balls speed is


The height of the ball as a function of time is given by \( h(t) = 20t - 4.9t^2 - 2 \). To find the ball's speed after 3 seconds, we need to calculate the derivative of the height function, \( h'(t) \), which gives the velocity of the ball at any time \( t \). The speed is the magnitude of the velocity.

Let's differentiate the height function:

\[
h(t) = 20t - 4.9t^2 - 2
\]

The derivative is:

\[
h'(t) = \frac{d}{dt} \left( 20t - 4.9t^2 - 2 \right) = 20 - 9.8t
\]

Now, substitute \( t = 3 \) to find the velocity at 3 seconds:

\[
h'(3) = 20 - 9.8 \times 3 = 20 - 29.4 = -9.4 \, \text{m/s}
\]

Since speed is the magnitude of velocity, the speed after 3 seconds is:

\[
\text{Speed} = | -9.4 | = 9.4 \, \text{m/s}
\]

The ball's speed after 3 seconds is **9.4 m/s**.

Would you like further details on this solution or have any other questions?

Here are five related questions:
1. What is the maximum height the ball reaches?
2. How long is the ball in the air before it hits the ground?
3. What is the velocity of the ball at its highest point?
4. How would the speed change if the ball were thrown with a different initial velocity?
5. How would the equation change if the ball were thrown from a different height?

**Tip**: The derivative of a position function gives the velocity, while the second derivative gives the acceleration.

A ball is thrown so that its height above the ground seconds after it has been thrown is given by the function h(t) = 20t - 4.9t^2 - 2. After 3 seconds, the ball's speed is

Explore how to calculate the speed of a ball after 3 seconds by differentiating the height function h(t) = 20t - 4.9t^2 - 2. This problem involves key calculus concepts such as differentiation and velocity, and is suitable for students in Grades 10-12.

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