https://file.aiquickdraw.com/mathbot/file/1729832557706bfmv9scj.jpg

Let's go through each part of the problem.

The height of the ball is given by the function:
\[
s(t) = 32 + 112t - 16t^2
\]
where \( s(t) \) is the height in feet at time \( t \) seconds.

### a. Determine the velocity \( v(t) \) of the ball after \( t \) seconds.
The velocity is the derivative of the height function \( s(t) \). So, let's differentiate \( s(t) \):
\[
v(t) = \frac{ds(t)}{dt} = 112 - 32t
\]
Thus, the velocity of the ball at time \( t \) is:
\[
v(t) = 112 - 32t
\]

### b. When does the ball reach its highest point?
The ball reaches its highest point when the velocity \( v(t) = 0 \) (the moment it stops moving upwards and begins to fall). Set \( v(t) = 0 \):
\[
112 - 32t = 0
\]
Solving for \( t \):
\[
t = \frac{112}{32} = 3.5 \text{ seconds}
\]
So, the ball reaches its highest point at \( t = 3.5 \) seconds.

### c. What is the height of the ball at the highest point?
To find the height at the highest point, substitute \( t = 3.5 \) into the original height function \( s(t) \):
\[
s(3.5) = 32 + 112(3.5) - 16(3.5)^2
\]
\[
s(3.5) = 32 + 392 - 16(12.25)
\]
\[
s(3.5) = 32 + 392 - 196 = 228 \text{ feet}
\]
Thus, the height of the ball at its highest point is 228 feet.

### d. When does the ball strike the ground?
The ball strikes the ground when \( s(t) = 0 \). Set the height equation equal to zero:
\[
32 + 112t - 16t^2 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -16 \), \( b = 112 \), and \( c = 32 \):
\[
t = \frac{-112 \pm \sqrt{112^2 - 4(-16)(32)}}{2(-16)}
\]
\[
t = \frac{-112 \pm \sqrt{12544 + 2048}}{-32}
\]
\[
t = \frac{-112 \pm \sqrt{14592}}{-32}
\]
\[
t = \frac{-112 \pm 120.87}{-32}
\]
This gives two solutions:
\[
t = \frac{-112 + 120.87}{-32} \approx \frac{8.87}{-32} \approx -0.277 \quad (\text{Not valid, negative time})
\]
\[
t = \frac{-112 - 120.87}{-32} \approx \frac{-232.87}{-32} \approx 7.28 \text{ seconds}
\]
So, the ball strikes the ground at approximately \( t = 7.28 \) seconds.

### e. With what velocity does the ball strike the ground?
To find the velocity when the ball strikes the ground, substitute \( t = 7.28 \) into the velocity function \( v(t) \):
\[
v(7.28) = 112 - 32(7.28)
\]
\[
v(7.28) = 112 - 232.96 = -120.96 \text{ ft/s}
\]
Thus, the velocity of the ball when it strikes the ground is approximately \(-120.96\) ft/s (negative because it's moving downward).

---

Would you like any further clarifications or more detailed explanations for any part of this? 

Here are 5 related questions you might want to explore:
1. How would the results change if the initial velocity were different?
2. What is the general formula for the time to reach the maximum height in projectile motion?
3. How does the gravitational constant \( g = 32 \, \text{ft/s}^2 \) affect the ball’s motion?
4. Can we use energy methods (kinetic and potential energy) to solve for the velocity at impact?
5. How would the problem change if the ball were thrown from a higher or lower initial height?

**Tip**: The negative sign in velocity indicates direction, so a negative velocity means the object is moving downward.

A ball is thrown upward from the roof of a 32-foot building with an initial velocity of 112 ft/s. The height after t seconds is given by the equation: s(t) = 32 + 112t - 16t^2. Find the following: a. The velocity after t seconds. b. When the ball reaches its highest point. c. The height of the ball at its highest point. d. When the ball strikes the ground. e. The velocity of the ball when it strikes the ground.

Learn how to solve a projectile motion problem where a ball is thrown from a 32-foot building with an initial velocity of 112 ft/s. This involves finding the velocity, maximum height, time to reach the highest point, and the velocity when the ball strikes the ground. Suitable for high school students studying physics or calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation