The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 21.5 m/s is 
h = 2 + 21.5t − 4.9t2
 after t seconds.
(a)
Find the velocity (in m/s) after 2 seconds and after 4 seconds.
v(2) = 
1.9
 
Correct: Your answer is correct.
 m/s
v(4) = 
-17.7
 
Correct: Your answer is correct.
 m/s
(b)
When does the projectile reach its maximum height? (Round your answer to two decimal places.)
2.19
 
Correct: Your answer is correct.
 s
(c)
What is the maximum height? (Round your answer to two decimal places.)
25.59
 
Incorrect: Your answer is incorrect.
 m
(d)
When does it hit the ground? (Round your answer to two decimal places.)

To tackle the problem step-by-step:

### Given:
The height of the projectile as a function of time is:

\[
h(t) = 2 + 21.5t - 4.9t^2
\]

Where:
- \( t \) is time in seconds,
- \( h(t) \) is height in meters.

### (c) Find the Maximum Height:
The maximum height occurs at the time when the projectile reaches its peak, i.e., at the time when the velocity is 0. The velocity \( v(t) \) is given by the derivative of the height function:

\[
v(t) = \frac{dh(t)}{dt} = 21.5 - 9.8t
\]

From part (b), the time at maximum height is \( t = 2.19 \) seconds. So, we substitute this into the height equation \( h(t) \):

\[
h(2.19) = 2 + 21.5(2.19) - 4.9(2.19)^2
\]

Let’s calculate it:

\[
h(2.19) = 2 + 21.5(2.19) - 4.9(4.7961)
\]
\[
h(2.19) = 2 + 47.785 - 23.5029
\]
\[
h(2.19) \approx 26.28 \, \text{m}
\]

So, the maximum height is approximately **26.28 meters**, not 25.59 m as previously calculated.

### (d) When Does the Projectile Hit the Ground?

To find when the projectile hits the ground, set \( h(t) = 0 \) and solve for \( t \):

\[
0 = 2 + 21.5t - 4.9t^2
\]

This is a quadratic equation. Rearranging:

\[
4.9t^2 - 21.5t - 2 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

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

Where \( a = 4.9 \), \( b = -21.5 \), and \( c = -2 \):

\[
t = \frac{-(-21.5) \pm \sqrt{(-21.5)^2 - 4(4.9)(-2)}}{2(4.9)}
\]
\[
t = \frac{21.5 \pm \sqrt{462.25 + 39.2}}{9.8}
\]
\[
t = \frac{21.5 \pm \sqrt{501.45}}{9.8}
\]
\[
t = \frac{21.5 \pm 22.39}{9.8}
\]

So, we have two possible solutions:

\[
t = \frac{21.5 + 22.39}{9.8} \quad \text{or} \quad t = \frac{21.5 - 22.39}{9.8}
\]
\[
t = \frac{43.89}{9.8} \quad \text{or} \quad t = \frac{-0.89}{9.8}
\]
\[
t \approx 4.48 \quad \text{or} \quad t \approx -0.09
\]

Since negative time doesn’t make sense in this context, the projectile hits the ground after approximately **4.48 seconds**.

---

Would you like me to explain any steps in more detail? Here are 5 related questions for further practice:

1. What is the velocity of the projectile at its maximum height?
2. If the initial velocity were increased, how would that affect the maximum height and time to hit the ground?
3. How would the equation change if the projectile were launched from 10 meters above ground instead of 2 meters?
4. Can you derive the formula for velocity from the height function?
5. What are the units of each term in the height equation \( h(t) = 2 + 21.5t - 4.9t^2 \)?

**Tip:** When solving quadratic equations, always check the physical relevance of your solutions, especially in time-dependent problems like this one!

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 21.5 m/s is h = 2 + 21.5t − 4.9t^2 after t seconds. (a) Find the velocity after 2 seconds and after 4 seconds. (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground?

Explore how to solve a projectile motion problem where the height is given by h(t) = 2 + 21.5t - 4.9t^2. This problem involves finding the velocity after 2 and 4 seconds, determining the time and value of maximum height, and solving for the time when the projectile hits the ground. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation