To solve the problem, we use the principles of projectile motion. The given data is:

• Initial velocity ($v_0$) = 20 m/s
• Angle of projection ($\theta$) = $55^\circ$
• Gravitational acceleration ($g$) = 9.8 m/s²
• Air resistance is negligible.

1. Maximum Height

The maximum height ($H$) is achieved when the vertical velocity component becomes zero. The formula is: $H = \frac{v_0^2 \sin^2\theta}{2g}$

First, calculate the vertical component of the velocity: $v_{y0} = v_0 \sin\theta = 20 \cdot \sin(55^\circ)$

$v_{y0} \approx 16.39 \, \text{m/s}$

Now, calculate $H$: $H = \frac{16.39^2}{2 \cdot 9.8} \approx \frac{268.69}{19.6} \approx 13.71 \, \text{m}$

Maximum Height: $H \approx 13.71 \, \text{m}$

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2. Time of Flight

The total time of flight ($T$) is calculated using the vertical motion. Since the object returns to the same height, the total time is twice the time to reach maximum height: $T = \frac{2 v_0 \sin\theta}{g}$

Substitute the values: $T = \frac{2 \cdot 20 \cdot \sin(55^\circ)}{9.8}$

$T = \frac{2 \cdot 16.39}{9.8} \approx \frac{32.78}{9.8} \approx 3.34 \, \text{s}$

Time of Flight: $T \approx 3.34 \, \text{s}$

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Final Results:

1. Maximum Height: $13.71 \, \text{m}$
2. Time of Flight: $3.34 \, \text{s}$

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Would you like more details or explanations? Here are some questions to consider next:

1. How do horizontal and vertical motions differ in projectile motion?
2. How does changing the angle of projection affect the time of flight and maximum height?
3. What is the horizontal range of the projectile in this scenario?
4. How would air resistance impact the maximum height and time of flight?
5. How do we calculate the initial velocity components mathematically?

Tip: When solving projectile motion problems, always decompose the velocity into horizontal and vertical components for clarity.