Question

Score! U OT pt

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

Step-by-step 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

$t=\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}$

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

$t=\frac{4\pm\sqrt{16-(-5600) } }{8}$

$t=\frac{4\pm\sqrt{16+5600 } }{8}$

$t=\frac{4\pm\sqrt{16+5616 } }{8}$

The discriminant $b^{2}-4ac>0$

so, there are two real roots.

$t=\frac{4\pm12\sqrt{39 } }{8}$

$t=\frac{4}{8}\pm\frac{12\sqrt{39 }}{8}$

$t=\frac{1}{2}\pm\frac{3\sqrt{39 }}{2}$

Use the positive square root to get a positive time.

$t=9.8675 s$

The pebble hits the ground after 9.8675 second