Question

What would be the equation of the cannonball’s path if it was launched with the same velocity, but from the ground instead?

Answer 1

The quadratic equation cannonball’s path if it was launched with the same velocity, but from the ground instead would be given by:

$h(t) = -4.9t^2 + 19.8t$

What is the quadratic equation for the height of a projectile?

Considering the gravity, it is modeled by:

$h(t) = -4.9t^2 + v_0t + h_0$

In which:

• $v_0$ is the initial velocity.

• $h_0$ is the initial height.

For this problem, we want:

• The same initial velocity, hence $v_0 = 19.8$.

• Launched from the ground, hence $h_0 = 0$.

Then, the equation is:

$h(t) = -4.9t^2 + 19.8t$

More can be learned about quadratic equations at brainly.com/question/24737967