Question

A waterfall has a height of 1400 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 16 feet per second. The height, h, of the pebble aftert
seconds is given by the equation hs - 167" + 16 + 1400. How long after the pebble is thrown will it hit the ground?

Answer 1

Answer:

2.9437 seconds

Step-by-step explanation:

I assume that you made a typo and the equation is -167s² + 16s + 1400

Find the zeros of the equation, or where it crosses the x-axis, to find at what time the pebble hits the ground. At these points, the height will be 0.

Use the quadratic equation

$\frac{-16 + \sqrt{16^{2} -4(-167)(1400)} }{2(-167)}$ = -2.8479

$\frac{-16 - \sqrt{16^{2} -4(-167)(1400)} }{2(-167)}$ = 2.9437

The negative answer doesn't make sense since time (s) starts from 0 and must be postive. Therefore, the answer is 2.9437