Question

Awaterfall has a height of 700 toet. A pebble is thrown upward from the top of the fals with an initial velocity of 20 foot per second. The height of the pebble h in foet anert seconds is given by the equation - - 168° + 204 + 700. How long after
the pebble is thrown will it hit the ground?
The time it takes for the pebble to hit the ground is about
(Simplify your answer. Type an integer or decimal rounded to the nearest benth as needed.)

Answer 1

Answer:

The time it takes for the pebble to hit the ground is about 7.3 seconds.

Step-by-step explanation:

Height after t seconds:

The height of the pebble after t seconds is given by:

$h(t) = -16t^2 + 20t + 700$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

How long after  the pebble is thrown will it hit the ground?

This is t for which $h(t) = 0$

So

$-16t^2 + 20t + 700 = 0$

Quadratic equation with $a = -16, b = 20, c = 700$

Then

$\Delta = 20^2 - 4(-16)(700) = 45200$

$t_{1} = \frac{-20 + \sqrt{45200}}{2(-16)} = -6$

$t_{2} = \frac{-20 - \sqrt{45200}}{2(-16)} = 7.3$

The time it takes for the pebble to hit the ground is about 7.3 seconds.