The height of the ball at the instant it is thrown corresponds to $t = 0$ (the time when the ball is released).

To find the height at $t = 0$, we substitute $t = 0$ into the given function for $h(t)$:

$h(0) = -4.9(0)^2 + 9.8(0) + 1$

Simplifying this expression:

$h(0) = 1$

Thus, the height of the ball at the instant it is thrown is 1 metre.

Would you like a more detailed explanation on how this model works or have any other questions?

Here are 5 related questions to expand on the concept:

1. What is the maximum height the ball reaches?
2. At what time does the ball hit the ground?
3. How does the initial velocity of the ball affect its trajectory?
4. How would the equation change if the ball was thrown from a higher or lower point?
5. What effect does changing the gravitational constant have on the ball's motion?

Tip: In a quadratic function like this one, the constant term (in this case, 1) often represents the initial height of the object when time $t = 0$.