To solve this problem, let's derive the formulas for the velocity $v(t)$ and height $h(t)$ of the ball based on the given information.

Given Data:

• Initial velocity: $v_0 = 51 \, \text{m/s}$
• Initial height: $h_0 = 9.3 \, \text{m}$
• Acceleration due to gravity: $a = -9.8 \, \text{m/s}^2$

Velocity $v(t)$:

The velocity of the ball as a function of time is given by the equation: $v(t) = v_0 + at$ Substituting the values: $v(t) = 51 - 9.8t$

Height $h(t)$:

The height of the ball as a function of time is given by: $h(t) = h_0 + v_0t + \frac{1}{2}at^2$ Substituting the values: $h(t) = 9.3 + 51t - 4.9t^2$

Final Equations:

1. Velocity: $v(t) = 51 - 9.8t \, \text{m/s}$
2. Height: $h(t) = 9.3 + 51t - 4.9t^2 \, \text{m}$

Let me know if you'd like to discuss these equations further or apply them to specific values of $t$.

Questions for Further Exploration:

1. What is the velocity of the ball after 3 seconds?
2. At what time does the ball reach its maximum height?
3. What is the maximum height the ball reaches?
4. When does the ball hit the ground?
5. How does changing the initial velocity affect the equations?

Tip: To find the time at which the ball reaches its maximum height, set $v(t) = 0$ and solve for $t$.