Let's go through each part of this projectile motion problem. Given data includes:

• Initial vertical velocity component, $v_{y0} = 20 \, \text{m/s}$
• Initial horizontal velocity component, $v_{x} = 25 \, \text{m/s}$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$ (acting downward)

a) Find the position and velocity of the ball after 2 s, 3 s, and 4 s.

For projectile motion, we can separate vertical and horizontal motion:

1. Horizontal position and velocity (constant):

   • Horizontal position: $x = v_x \cdot t$
   • Horizontal velocity remains constant: $v_x = 25 \, \text{m/s}$

2. Vertical position and velocity (under gravity):

   • Vertical position: $y = v_{y0} \cdot t - \frac{1}{2} g \cdot t^2$
   • Vertical velocity: $v_y = v_{y0} - g \cdot t$

Let's calculate these at $t = 2 \, \text{s}$, $t = 3 \, \text{s}$, and $t = 4 \, \text{s}$:

1. At $t = 2 \, \text{s}$:

   • $x = 25 \cdot 2 = 50 \, \text{m}$
   • $y = 20 \cdot 2 - \frac{1}{2} \cdot 9.8 \cdot 2^2 = 40 - 19.6 = 20.4 \, \text{m}$
   • $v_y = 20 - 9.8 \cdot 2 = 20 - 19.6 = 0.4 \, \text{m/s}$ (upward)

2. At $t = 3 \, \text{s}$:

   • $x = 25 \cdot 3 = 75 \, \text{m}$
   • $y = 20 \cdot 3 - \frac{1}{2} \cdot 9.8 \cdot 3^2 = 60 - 44.1 = 15.9 \, \text{m}$
   • $v_y = 20 - 9.8 \cdot 3 = 20 - 29.4 = -9.4 \, \text{m/s}$ (downward)

3. At $t = 4 \, \text{s}$:

   • $x = 25 \cdot 4 = 100 \, \text{m}$
   • $y = 20 \cdot 4 - \frac{1}{2} \cdot 9.8 \cdot 4^2 = 80 - 78.4 = 1.6 \, \text{m}$
   • $v_y = 20 - 9.8 \cdot 4 = 20 - 39.2 = -19.2 \, \text{m/s}$ (downward)

b) How much time is required to reach the highest point of the trajectory?

At the highest point, vertical velocity $v_y = 0$.

Using $v_y = v_{y0} - g \cdot t$: $0 = 20 - 9.8 \cdot t \Rightarrow t = \frac{20}{9.8} \approx 2.04 \, \text{s}$

c) How high is this point?

Using $y = v_{y0} \cdot t - \frac{1}{2} g \cdot t^2$ at $t = 2.04 \, \text{s}$: $y = 20 \cdot 2.04 - \frac{1}{2} \cdot 9.8 \cdot (2.04)^2$ $y \approx 20.4 \, \text{m}$

d) How much time (after launch) is required for the ball to return to its original level?

The total time of flight $T$ for a projectile launched from and landing at the same height is given by $T = 2 \times t_{\text{highest}}$: $T = 2 \times 2.04 \approx 4.08 \, \text{s}$

e) How far has it traveled horizontally during this time?

Using $x = v_x \cdot T$: $x = 25 \cdot 4.08 \approx 102 \, \text{m}$

Would you like further details or have any specific questions? Here are some related questions:

1. How does gravity affect the vertical and horizontal components of the projectile’s motion?
2. What happens to the total energy of the projectile at the highest point?
3. How would changing the angle of projection affect the time of flight?
4. If the initial speed were doubled, how would the maximum height and range be affected?
5. Why is horizontal velocity constant in projectile motion without air resistance?

Tip: Remember, in projectile motion, horizontal and vertical motions are independent of each other.