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.