https://file.aiquickdraw.com/mathbot/file/1729860979625yw9lcmxv.jpg

Let's go through each question from the image and solve them step by step.

### Question 1:
**A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed?**

We can use the kinematic equation that relates the final velocity (which is 0 m/s at the top of the jump), the initial velocity (takeoff speed), acceleration (due to gravity), and displacement (height of the jump):
\[
v^2 = u^2 + 2as
\]
Where:
- \(v = 0\) m/s (velocity at the highest point),
- \(u\) is the takeoff speed (what we need to find),
- \(a = -9.8\) m/s² (acceleration due to gravity, negative because it opposes motion),
- \(s = 2.0\) m (the height reached).

Rearranging the equation for \(u\):
\[
u = \sqrt{2as}
\]
\[
u = \sqrt{2 \times 9.8 \times 2} = \sqrt{39.2} \approx 6.26 \, \text{m/s}
\]
So, the closest answer is **6.3 m/s**.

---

### Question 2:
**A cannonball is fired straight up at a speed of 25 m/s. What is the maximum altitude that it will reach?**

We use the same kinematic equation as before:
\[
v^2 = u^2 + 2as
\]
Where:
- \(v = 0\) m/s (velocity at the highest point),
- \(u = 25\) m/s (initial velocity),
- \(a = -9.8\) m/s²,
- \(s\) is the maximum height (what we need to find).

Rearranging for \(s\):
\[
s = \frac{u^2 - v^2}{2a}
\]
\[
s = \frac{25^2 - 0^2}{2 \times 9.8} = \frac{625}{19.6} \approx 31.89 \, \text{m}
\]
The closest answer is **31.9 m**.

---

### Question 3:
**How long will it take the cannonball in problem #2 to reach its maximum height?**

We can use the kinematic equation for velocity:
\[
v = u + at
\]
Where:
- \(v = 0\) m/s (velocity at the highest point),
- \(u = 25\) m/s (initial velocity),
- \(a = -9.8\) m/s²,
- \(t\) is the time to reach maximum height.

Rearranging for \(t\):
\[
t = \frac{v - u}{a} = \frac{0 - 25}{-9.8} \approx 2.55 \, \text{seconds}
\]
The answer is **2.55 s**.

---

### Question 4:
**At what velocity (magnitude and direction) would you hit the floor if you stepped off a chair 0.75 m high?**

We use the kinematic equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \(v\) is the final velocity (what we need to find),
- \(u = 0\) m/s (initial velocity),
- \(a = 9.8\) m/s² (since gravity is acting downward),
- \(s = 0.75\) m.

Rearranging for \(v\):
\[
v = \sqrt{u^2 + 2as} = \sqrt{0 + 2 \times 9.8 \times 0.75} = \sqrt{14.7} \approx 3.83 \, \text{m/s}
\]
The direction is downward, so the answer is **-3.83 m/s**.

---

Would you like further details on any of the questions? Here are 5 related questions you could explore:

1. What factors influence the time of flight for a projectile?
2. How does gravity affect the maximum height a projectile can reach?
3. How would air resistance alter the results for the kangaroo or cannonball?
4. Can you derive the equation for a falling object from Newton’s laws?
5. How would the initial speed of the cannonball affect its total flight time?

**Tip:** When dealing with kinematic problems, always list out known variables first and choose the appropriate equation.

The image contains four physics questions, dealing with takeoff speed, maximum altitude, time to maximum height, and impact velocity.

Explore solutions to kinematics problems covering takeoff speed of a kangaroo, maximum height of a cannonball, time to peak height, and impact velocity after a fall. These step-by-step solutions include key physics formulas for high school students.

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