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We are given the height function \( H(t) = -16t^2 + 40t + 24 \), where \( H(t) \) represents the height of the ball above the ground at time \( t \) in seconds. Let's address each part of the problem.

### Part (a): Find the velocity of the ball after two seconds.

Velocity is the derivative of the height function with respect to time, \( v(t) = H'(t) \).

1. First, find the derivative of \( H(t) \):
   \[
   H'(t) = \frac{d}{dt}(-16t^2 + 40t + 24) = -32t + 40
   \]

2. Now, evaluate the velocity at \( t = 2 \) seconds:
   \[
   v(2) = -32(2) + 40 = -64 + 40 = -24 \, \text{feet per second}
   \]
   So, the velocity of the ball after two seconds is \(-24\) feet per second, meaning it's falling downwards.

---

### Part (b): When does the ball hit the ground, and what is its impact velocity?

The ball hits the ground when \( H(t) = 0 \).

1. Set the height equation equal to zero:
   \[
   -16t^2 + 40t + 24 = 0
   \]

2. Solve this quadratic equation using the quadratic formula:
   \[
   t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For \( a = -16 \), \( b = 40 \), and \( c = 24 \):
   \[
   t = \frac{-40 \pm \sqrt{40^2 - 4(-16)(24)}}{2(-16)} = \frac{-40 \pm \sqrt{1600 + 1536}}{-32} = \frac{-40 \pm \sqrt{3136}}{-32} = \frac{-40 \pm 56}{-32}
   \]

   This gives two solutions:
   \[
   t = \frac{-40 + 56}{-32} = \frac{16}{-32} = -0.5 \, (\text{discard, negative time doesn't make sense})
   \]
   \[
   t = \frac{-40 - 56}{-32} = \frac{-96}{-32} = 3 \, \text{seconds}
   \]
   So, the ball hits the ground after 3 seconds.

3. To find the impact velocity, evaluate \( v(t) \) at \( t = 3 \):
   \[
   v(3) = -32(3) + 40 = -96 + 40 = -56 \, \text{feet per second}
   \]
   The impact velocity is \(-56\) feet per second (downward).

---

### Part (c): When does the ball have a velocity of zero? What physical interpretation should be given to this time?

1. Set \( v(t) = 0 \) to find when the velocity is zero:
   \[
   -32t + 40 = 0 \quad \Rightarrow \quad t = \frac{40}{32} = 1.25 \, \text{seconds}
   \]

2. The ball has a velocity of zero at \( t = 1.25 \) seconds. This is the moment when the ball reaches its maximum height (the turning point from moving upwards to downwards).

---

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. What is the maximum height of the ball?
2. How would the equations change if the ball was thrown upwards with a different initial velocity?
3. How does air resistance affect the ball's motion in real-life scenarios?
4. What would happen to the ball's motion if it was dropped from rest instead of being thrown?
5. How can we determine the total distance traveled by the ball until it hits the ground?

**Tip:** Remember that velocity is the rate of change of position (height), and when velocity is zero, it typically indicates a turning point in the motion.

Given the height function H(t) = -16t^2 + 40t + 24, find the velocity of the ball after two seconds, when the ball hits the ground and its impact velocity, and when the ball's velocity is zero.

Learn how to solve a physics problem involving the height of a ball thrown from a cliff using the function H(t) = -16t^2 + 40t + 24. This problem covers key concepts in calculus such as derivatives and solving quadratic equations. Suitable for Grades 10-12.

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