To determine the angle at which Fred must make each of his jumps, we can break this problem into horizontal and vertical motion components. Here's the detailed solution:

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1. Identify Given Information:

• Horizontal distance between lily pads ($x$): $2.4 \, \text{m}$
• Jump speed ($v$): $5.0 \, \text{m/s}$
• Time to jump ($t$): $0.60 \, \text{s}$
• Acceleration due to gravity ($g$): $10.0 \, \text{m/s}^2$

2. Set Up Equations:

Fred's jump follows a parabolic trajectory. The horizontal and vertical motion are analyzed separately:

(a) Horizontal motion:

$x = v_x \cdot t$ where $v_x = v \cos \theta$. Substituting: $2.4 = (5.0 \cos \theta) \cdot 0.60$ $\cos \theta = \frac{2.4}{5.0 \cdot 0.60}$ $\cos \theta = 0.8$

(b) Vertical motion:

In vertical motion, the frog must return to the same height after $t = 0.60 \, \text{s}$. The equation for vertical displacement is: $y = v_y t - \frac{1}{2} g t^2$ Since $y = 0$ at both the start and end of the jump: $0 = (v \sin \theta) \cdot 0.60 - \frac{1}{2} (10.0) (0.60)^2$ Simplify: $0 = 3.0 \sin \theta - 1.8$ $\sin \theta = \frac{1.8}{3.0}$ $\sin \theta = 0.6$

3. Solve for $\theta$:

Now we have: $\cos \theta = 0.8, \quad \sin \theta = 0.6$

The angle $\theta$ satisfies: $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{0.6}{0.8}$ $\theta = \tan^{-1}(0.75)$ $\theta \approx 36.9^\circ$

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Final Answer:

Fred must make his jumps at an angle of approximately $36.9^\circ$.

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Would you like a more detailed breakdown or have further questions?

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Follow-Up Questions:

1. How would the angle change if the lily pads were spaced further apart?
2. What happens if Fred’s jump speed decreases?
3. How does gravity ($g$) influence Fred’s trajectory?
4. What is the maximum height Fred reaches during his jump?
5. Can Fred make the jump if he changes his angle but keeps the same speed?

Tip:

When solving projectile motion problems, always separate horizontal and vertical components for clarity.