Question

A waterfall has a height of 900 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 24 feet per second. The​ height, h, of the pebble after t seconds is given by the equation h equals negative 16 t squared plus 24 t plus 900. How long after the pebble is thrown will it hit the​ ground?

Answer 1

Answer:

The pebble will hit the ground 8.29 second after it is thrown.

Step-by-step explanation:

Given that,

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

h= -16t²+24 t+900.

When the pebble hit the ground, the height of the pebble will be zero.

So putting h=0 in the given equation.

0 = -16t²+24 t+900

⇒-4(4t²-6t-225)=0

⇒4t²-6t-225=0

Applying quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, here a=4, b= -6 and c= -225.

$\therefore t=\frac{-(-6)\pm\sqrt{(-6)^2-4.4.(-225)}}{2.4}$

⇒ t = 8.29, -6.79

Since time can not negative.

∴t=8.29 seconds.

The pebble will hit the ground 8.29 second after it is thrown.