Question

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.

Answer 1

`\\ \tt \longmapsto \: h = - 4.8 {t}^(2) + 8.2t + 1.8 \\ \\ \tt \longmapsto \: h = - 4.8(1.7) {}^(2) + 8.2 * 1.7 + 1.8 \\ \\ \tt \longmapsto \: - 4.8 * 2.89 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 17.06`