A ball is thrown and it's height at time t seconds is given by h(t) = 100 + 40t — 5t2. Find
(i) initial height
(ii) maximum height
(iii) velocity after 6 seconds
(iv) speed after 6 seconds
(v) velocity when the ball hits the ground
(vi) ball's acceleration

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).

---

### 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.

Explore detailed solutions for a projectile motion problem involving initial height, maximum height, velocities, speeds, and acceleration of a thrown ball. Learn how to calculate these parameters step-by-step with practical examples.

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