Q:

# 8. John launches a toy rocket into the air from the ground. The rocket is in the air for 10s before returning to earth. Jeans that after 2 s the rocket has a height of 20 m in the air. 14 a) Give the zeros and one point on the rocket&#39;s path. You can include a diagram. (1) b) Use your values in a) to find the equation of the path. (3) Determine the equation of the parabola with zeros 6 and - 2 and an intercept of 9. Give your answer the factorized form AND the general form.

Accepted Solution

A:
a) We can represent the rocket's height above the ground as a function of time using a quadratic equation, since the path of the rocket will form a parabolic shape. Let's denote the time that the rocket spends in the air as t, and the maximum height reached by the rocket as h. From the problem statement, we know that the rocket is in the air for 10 seconds, so t = 10. We also know that after 2 seconds, the rocket is at a height of 20 meters, so h = 20. We can use this information to form the following equation for the height of the rocket as a function of time: h(t) = a(t - 2)(t - 10) where a is a constant that determines the shape of the parabola. To solve for a, we can use the fact that the maximum height of the rocket is 20 meters, which occurs at the vertex of the parabola. The x-coordinate of the vertex is given by the average of the zeros, which is: (-2 + 6) / 2 = 2 So the maximum height occurs at t = 2, which means that: h(2) = a(2 - 2)(2 - 10) = 0 Since h(2) = 20, we can solve for a: 20 = a(0) a = undefined This means that the quadratic equation that we formed is not valid for this situation, and we cannot represent the rocket's path as a parabolic shape. b) Since we were unable to find the quadratic equation that represents the rocket's path, we cannot answer this part of the question.