Question

A waterfall has a height of 1100 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 20 feet per second. The height of the pebble h in feet after t seconds is given by the equation h = -16 t^2 + 20*t + 1100. How long after the pebble is thrown will it hit the​ ground?

Answer 1

8.94 seconds

Explanation:

The ground is at height 0.

We set the equation equal to zero and solve for t.

h = -16 t^2 + 20*t + 1100

We want h = 0, so we get

-16 t^2 + 20*t + 1100 = 0

4t^2 - 5t - 275 = 0

`t = (-b \pm √(b^2 - 4ac))/(2a)`

`t = (-(-5) \pm √((-5)^2 - 4(4)(-275)))/(2(4))`

`t = (5 \pm √(25 + 4400))/(8)`

`t = (5 \pm √(4425))/(8)`

`t = (5 \pm 5√(177))/(8)`

`t = 8.94 [\tex]   or` t = -7.69 [/tex]

We discard the negative answer.

Answer: 8.94 seconds