This problem involves projectile motion. Let's break it down and solve each part step by step.

Given:

• Initial velocity ($v_0$) = 100 m/s
• Angle of projection ($\theta$) = 60 degrees
• Gravitational acceleration ($g$) = 9.8 m/s²

Part a: Find the velocity of the ball after 2 seconds

To calculate the velocity after 2 seconds, we need to find both the horizontal and vertical components of the velocity.

1. Horizontal velocity ($v_x$): $v_x = v_0 \cos(\theta) = 100 \times \cos(60^\circ) = 100 \times 0.5 = 50 \, \text{m/s}$ Horizontal velocity remains constant since there is no horizontal acceleration (ignoring air resistance).

2. Vertical velocity ($v_y$) after time $t$: $v_y = v_0 \sin(\theta) - g t$ $v_y = 100 \times \sin(60^\circ) - 9.8 \times 2 = 100 \times 0.866 - 19.6 = 86.6 - 19.6 = 67 \, \text{m/s}$

The total velocity $v$ is found using the Pythagorean theorem: $v = \sqrt{v_x^2 + v_y^2}$ $v = \sqrt{50^2 + 67^2} = \sqrt{2500 + 4489} = \sqrt{6989} \approx 83.6 \, \text{m/s}$

So, the velocity after 2 seconds is approximately 83.6 m/s.

Part b: Maximum height of the ball

The maximum height occurs when the vertical velocity becomes zero. Using the vertical motion formula: $v_y = v_0 \sin(\theta) - g t$ At the maximum height, $v_y = 0$, so: $0 = v_0 \sin(\theta) - g t$ $t = \frac{v_0 \sin(\theta)}{g} = \frac{100 \times 0.866}{9.8} = \frac{86.6}{9.8} \approx 8.84 \, \text{seconds} (time to reach maximum height)$

Now, using the vertical displacement formula to find the maximum height: $h_{\text{max}} = v_0 \sin(\theta) \times t - \frac{1}{2} g t^2$ $h_{\text{max}} = 100 \times 0.866 \times 8.84 - \frac{1}{2} \times 9.8 \times (8.84)^2$ $h_{\text{max}} \approx 766.024 - 381.07 = 384.95 \, \text{meters}$

So, the maximum height is approximately 385 meters.

Part c: Range of the ball

The range of a projectile is given by: $R = \frac{v_0^2 \sin(2\theta)}{g}$ $R = \frac{100^2 \sin(120^\circ)}{9.8} = \frac{10000 \times 0.866}{9.8} = \frac{8660}{9.8} \approx 883.7 \, \text{meters}$

So, the range is approximately 883.7 meters.

Part d: Time of flight

The time of flight is twice the time taken to reach the maximum height: $t_{\text{flight}} = 2 \times t_{\text{max}} = 2 \times 8.84 = 17.68 \, \text{seconds}$

So, the total time of flight is approximately 17.68 seconds.

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Summary:

• Velocity after 2 seconds: 83.6 m/s
• Maximum height: 385 meters
• Range: 883.7 meters
• Time of flight: 17.68 seconds

Would you like more details on any part? Here are five related questions to extend this topic:

1. How would air resistance affect the range and maximum height?
2. Can you calculate the time when the ball reaches half of the maximum height?
3. What would the velocity be after 5 seconds?
4. How would the results change if the angle of projection were 45 degrees?
5. What is the impact of gravity variation on the range?

Tip: Remember, the horizontal velocity component remains constant in the absence of air resistance, while the vertical component is affected by gravity.