To find the height $H$ of the cannonball at specific times using the given formula:

$H = 60t - 5t^2$

where $t$ is the time in seconds, we substitute the given time values into the equation.

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i. After 1 second ($t = 1$):

Substitute $t = 1$ into $H = 60t - 5t^2$:

$H = 60(1) - 5(1)^2$ $H = 60 - 5 = 55 \, \text{meters}$

The height of the cannonball after 1 second is $55 \, \text{meters}$.

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ii. After 3 seconds ($t = 3$):

Substitute $t = 3$ into $H = 60t - 5t^2$:

$H = 60(3) - 5(3)^2$ $H = 180 - 5(9)$ $H = 180 - 45 = 135 \, \text{meters}$

The height of the cannonball after 3 seconds is $135 \, \text{meters}$.

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Final Answer:

1. After 1 second: $55 \, \text{meters}$
2. After 3 seconds: $135 \, \text{meters}$

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Do you want further explanations on how the height changes over time or need a graph of the equation?

Here are 5 related questions to expand this concept:

1. How long does it take for the cannonball to reach its maximum height?
2. What is the maximum height reached by the cannonball?
3. How long does it take for the cannonball to hit the ground?
4. Plot the graph of $H = 60t - 5t^2$.
5. What is the height of the cannonball at $t = 5$ seconds?

Tip: The term $-5t^2$ indicates a parabolic motion, and the height decreases after reaching its maximum.