Question

Score! U OT pt

1 of 10 ( complete)
IVY JUICU IU, U UI TUPU
ses
32.6.1
Skill Builder
5 Question Help
ome
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 after t
seconds is given by the equation h - 16t" + 16 + 1400. How long after the pebble is thrown will it hit the ground?
The pebble will hit the ground about
seconds after it is thrown.

Answer 1

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 after t seconds is given by the equation h equals negative 16 t squared plus 16 t plus 1400

h=−16t2+16t+1400. How long after the pebble is thrown will it hit the​ ground?

Answer

The pebble hits the ground after 9.8675 s

Explanation:

Given

waterfall height = 1400 feet

initial velocity = 16 feet per second

The height, h, of the pebble after t seconds is given by the equation.

`h(t) = -16t^(2)+16t+1400`

The pebble hits the ground when
`h = 0`

`h=-16t^(2)+16t+1400` ---------------(1)

put
`h=0` in equation (1)

`0=-16t^(2)+16t+1400`

`-16t^(2)+16t+1400=0`

Divide by -4 to simplify this equation

`4t^(2)-4t-350=0`

using the Quadratic Formula where

a = 4, b = -4, and c = -350

`t=\frac{-b\pm\sqrt{b^(2)-4ac } }{2a}`

`t=\frac{-(-4)\pm\sqrt{(-4)^(2)-4(4)(-350) } }{2(4)}`

`t=(4\pm√(16-(-5600) ) )/(8)`

`t=(4\pm√(16+5600 ) )/(8)`

`t=(4\pm√(16+5616 ) )/(8)`

The discriminant
`b^(2)-4ac>0`

so, there are two real roots.

`t=(4\pm12√(39 ) )/(8)`

`t=(4)/(8)\pm(12√(39 ))/(8)`

`t=(1)/(2)\pm(3√(39 ))/(2)`

Use the positive square root to get a positive time.

`t=9.8675 s`

The pebble hits the ground after 9.8675 second