The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 23.5 m/s is 
h = 3 + 23.5t − 4.9t2
 after t seconds.
(e)
With what velocity (in m/s) does it hit the ground? (Round your answer to two decimal places.)
24.80
 

To determine the velocity at which the projectile hits the ground, we first need to calculate the time \( t \) at which it reaches the ground. The height \( h \) is given by the equation:

\[
h = 3 + 23.5t - 4.9t^2
\]

The projectile hits the ground when \( h = 0 \). Setting \( h \) to 0 and solving for \( t \):

\[
0 = 3 + 23.5t - 4.9t^2
\]

Rearranging the equation:

\[
4.9t^2 - 23.5t - 3 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 4.9 \), \( b = -23.5 \), and \( c = -3 \).

Let's substitute the values:

\[
t = \frac{-(-23.5) \pm \sqrt{(-23.5)^2 - 4(4.9)(-3)}}{2(4.9)}
\]

Simplifying the expression under the square root:

\[
t = \frac{23.5 \pm \sqrt{23.5^2 + 4(4.9)(3)}}{9.8}
\]
\[
t = \frac{23.5 \pm \sqrt{552.25 + 58.8}}{9.8}
\]
\[
t = \frac{23.5 \pm \sqrt{611.05}}{9.8}
\]
\[
t = \frac{23.5 \pm 24.72}{9.8}
\]

Now, we calculate the two possible values for \( t \):

1. \( t_1 = \frac{23.5 + 24.72}{9.8} \approx \frac{48.22}{9.8} \approx 4.92 \, \text{seconds} \)
2. \( t_2 = \frac{23.5 - 24.72}{9.8} \approx \frac{-1.22}{9.8} \approx -0.12 \, \text{seconds} \) (which is not physically meaningful, so we discard this value).

The time \( t = 4.92 \, \text{seconds} \) is the moment the projectile hits the ground.

### Step 2: Find the velocity at \( t = 4.92 \, \text{seconds} \)

The velocity at any time \( t \) is given by the derivative of the height function \( h(t) \). Differentiating \( h(t) = 3 + 23.5t - 4.9t^2 \) with respect to \( t \):

\[
v(t) = \frac{d}{dt}[3 + 23.5t - 4.9t^2] = 23.5 - 9.8t
\]

Substitute \( t = 4.92 \) into the velocity equation:

\[
v(4.92) = 23.5 - 9.8(4.92)
\]
\[
v(4.92) = 23.5 - 48.22
\]
\[
v(4.92) = -24.72 \, \text{m/s}
\]

Thus, the velocity with which the projectile hits the ground is approximately **−24.72 m/s**. The negative sign indicates that the projectile is moving downward.

Would you like more details on any part of this process?

### Related Questions:
1. How do you derive the velocity function from the height function?
2. What are other methods to find the velocity besides differentiation?
3. Can we use energy principles to find the impact velocity?
4. What would the total time of flight be if the projectile were launched from ground level?
5. How does air resistance affect the calculations in real-life scenarios?

### Tip:
Always double-check if your time \( t \) is physically meaningful before using it in further calculations.

With what velocity (in m/s) does a projectile hit the ground if its height is given by h = 3 + 23.5t − 4.9t^2 and it is shot vertically upward from 3 m above ground with an initial velocity of 23.5 m/s? (Round your answer to two decimal places.)

Learn how to calculate the velocity of a projectile when it hits the ground, using quadratic equations and differentiation. This detailed step-by-step solution covers key concepts in algebra and kinematics, suitable for Grades 10-12.

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