A ball is kicked off a pier and lands in the sea. The balls height T seconds after it was kicked is given by the equation h(T)=2/7x(T+3)(7-T) after how many seconds is the ball at the same height as the pier again Height of pier is 6

Given the height equation:

\[ h(T) = \frac{2}{7} (T + 3)(7 - T) \]

We need to determine the time \( T \) when the ball is at the same height as the pier, which is 6 units.

Set the height equation equal to 6 and solve for \( T \):

\[ 6 = \frac{2}{7} (T + 3)(7 - T) \]

Multiply both sides by 7 to clear the fraction:

\[ 42 = 2 (T + 3)(7 - T) \]

Divide both sides by 2:

\[ 21 = (T + 3)(7 - T) \]

Expand the right-hand side:

\[ 21 = 7T - T^2 + 21 - 3T \]

Combine like terms:

\[ 21 = -T^2 + 4T + 21 \]

Subtract 21 from both sides:

\[ 0 = -T^2 + 4T \]

Factor out \(-T\):

\[ 0 = -T(T - 4) \]

Set each factor equal to zero:

1. \[ -T = 0 \]
   \[ T = 0 \]

2. \[ T - 4 = 0 \]
   \[ T = 4 \]

Therefore, the ball is at the same height as the pier (6 units) at \( T = 0 \) seconds (initially) and \( T = 4 \) seconds.

Would you like more details on this solution or have any other questions?

Here are 5 more questions you might ask:

1. How do you find the maximum height the ball reaches?
2. What is the significance of the coefficients in the height equation?
3. How does the shape of the trajectory change with different coefficients?
4. What is the time interval during which the ball is above the pier?
5. How do you determine the time of flight of the ball?

**Tip:** When solving quadratic equations, remember to look for opportunities to factor the equation after setting it to zero. This can simplify the process of finding the solutions.

Learn how to solve the equation h(T) = (2/7)(T + 3)(7 - T) to find when the ball's height matches the pier's height (6 units). This problem involves quadratic equations, factoring, and solving equations, suitable for high school students.

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