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7nadin3 [17]
3 years ago
12

Score! U OT pt

Mathematics
1 answer:
Setler79 [48]3 years ago
7 0

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

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