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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.

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.

The pebble hits the ground when  

---------------(1)

put in equation (1)

Divide by -4 to simplify this equation

using the Quadratic Formula where

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

The discriminant

so, there are two real roots.

Use the positive square root to get a positive time.

The pebble hits the ground after 9.8675 second

y = 0.77(x)

step-by-step explanation:

the jacket retains this value after 23% cut or retains 77% of original value. plug original 87 jacket price for x and the remaining value should be y.