Question

A rocket is launched from atop a 55-foot cliff with an initial velocity of 138 ft/s. a. Substitute the values into the vertical motion formula h = −16t2 + vt + c. Let h = 0. b. Use the quadratic formula find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.

Answer 1

It would take 9 seconds.

Substituting the values into the equation, we have:
 
0 = -16t² + 138t + 55

The quadratic formula is:
$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Inserting our information we have:
$t=\frac{-138\pm \sqrt{138^2-4(-16)(55)}}{2(-16)} \\ \\=\frac{-138\pm \sqrt{19044--3520}}{-32} \\ \\=\frac{-138\pm \sqrt{19044+3520}}{-32} \\ \\=\frac{-138\pm \sqrt{22564}}{-32} \\ \\=\frac{-138\pm 150}{-32}=\frac{-138+150}{-32}\text{ or } \frac{-138-150}{-32} \\ \\=\frac{12}{-32}\text { or } \frac{288}{32}=-0.375 \text{ or } 9$