To determine the maximum height the ball will reach, we can break down its initial velocity into vertical and horizontal components and then focus on the vertical motion.

Step-by-Step Solution

1. Calculate the vertical component of the initial velocity: Given:

   • Initial velocity $v = 30 \, \text{m/s}$
   • Angle of projection $\theta = 33^\circ$

   The vertical component of the velocity $v_y$ is: $v_y = v \sin \theta = 30 \sin(33^\circ)$

   Calculating this: $v_y \approx 30 \times 0.5446 = 16.338 \, \text{m/s}$

2. Calculate the maximum height $h$: At the highest point, the vertical velocity becomes zero due to gravity. Using the kinematic equation: $v_y^2 = u_y^2 - 2gh$ where:

   • $v_y = 0 \, \text{m/s}$ (final vertical velocity at the maximum height)
   • $u_y = 16.338 \, \text{m/s}$ (initial vertical velocity)
   • $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)

   Rearranging for $h$: $h = \frac{u_y^2}{2g} = \frac{(16.338)^2}{2 \times 9.8}$

   Calculating this: $h = \frac{267.001}{19.6} \approx 13.62 \, \text{meters}$

Final Answer:

The maximum height the ball will reach is approximately 13.62 meters.

Would you like details on any steps or have further questions?

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Here are five related questions for deeper understanding:

1. How long will the ball remain in the air?
2. What is the range (horizontal distance) of the ball?
3. How would the maximum height change if the angle were increased to 45°?
4. What effect would doubling the initial speed have on the maximum height?
5. How does gravity influence the time taken to reach the maximum height?

Tip: Breaking down projectile motion into vertical and horizontal components simplifies calculations and helps analyze each part separately.