Let's solve each part of the problem step by step.

Given the height function: $h(t) = 100 + 40t - 5t^2$ where $h(t)$ represents the height in meters at time $t$ seconds.

(i) Initial Height

The initial height is the height when $t = 0$. $h(0) = 100 + 40(0) - 5(0)^2 = 100 \text{ meters}$ So, the initial height is 100 meters.

(ii) Maximum Height

The maximum height is reached when the velocity is zero. The velocity function is the derivative of the height function with respect to time $t$: $v(t) = \frac{dh(t)}{dt} = 40 - 10t$ Set $v(t) = 0$ to find the time at which the maximum height occurs: $40 - 10t = 0$ $t = \frac{40}{10} = 4 \text{ seconds}$ Now, substitute $t = 4$ into the height function to find the maximum height: $h(4) = 100 + 40(4) - 5(4)^2 = 100 + 160 - 80 = 180 \text{ meters}$ So, the maximum height is 180 meters.

(iii) Velocity After 6 Seconds

The velocity after 6 seconds can be found by substituting $t = 6$ into the velocity function: $v(6) = 40 - 10(6) = 40 - 60 = -20 \text{ m/s}$ So, the velocity after 6 seconds is -20 m/s (negative indicates downward direction).

(iv) Speed After 6 Seconds

Speed is the magnitude of velocity. Since the velocity is $-20$ m/s: $\text{Speed} = |v(6)| = 20 \text{ m/s}$ So, the speed after 6 seconds is 20 m/s.

(v) Velocity When the Ball Hits the Ground

The ball hits the ground when $h(t) = 0$. Solve for $t$: $100 + 40t - 5t^2 = 0$ This is a quadratic equation: $5t^2 - 40t - 100 = 0$ Divide the entire equation by 5: $t^2 - 8t - 20 = 0$ Solve using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -8$, and $c = -20$: $t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-20)}}{2(1)}$ $t = \frac{8 \pm \sqrt{64 + 80}}{2}$ $t = \frac{8 \pm \sqrt{144}}{2}$ $t = \frac{8 \pm 12}{2}$ The two possible values for $t$ are: $t = \frac{20}{2} = 10 \text{ seconds} \quad \text{(valid time when the ball hits the ground)}$ $t = \frac{-4}{2} = -2 \text{ seconds} \quad \text{(not physically meaningful)}$ So, the ball hits the ground at $t = 10$ seconds. Now, find the velocity at that time: $v(10) = 40 - 10(10) = 40 - 100 = -60 \text{ m/s}$ So, the velocity when the ball hits the ground is -60 m/s.

(vi) Ball's Acceleration

The acceleration is the derivative of the velocity function. Since the velocity function is $v(t) = 40 - 10t$, the acceleration is: $a(t) = \frac{dv(t)}{dt} = -10 \text{ m/s}^2$ So, the ball's acceleration is -10 m/s² (constant and downward).

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Summary of Answers:

1. Initial height: 100 meters
2. Maximum height: 180 meters
3. Velocity after 6 seconds: -20 m/s
4. Speed after 6 seconds: 20 m/s
5. Velocity when the ball hits the ground: -60 m/s
6. Ball's acceleration: -10 m/s²

Would you like further details or have any questions about these results? Here are some related questions to explore:

1. How long does it take for the ball to reach half of its maximum height?
2. What is the height of the ball after 3 seconds?
3. How would the results change if the ball were thrown with a different initial velocity?
4. What would be the maximum height if the initial velocity were doubled?
5. How would the acceleration change if air resistance were considered?

Tip: The sign of the velocity indicates the direction of the motion—positive for upward and negative for downward.