Question

A rock is thrown in the air from the edge of a seaside cliff. Its height in feet is represented by f(x) = –16(x2 – 3x – 18), where x is the number of seconds since the rock was thrown. The height of the rock is 0 feet when it hits the water.How long does it take the rock to hit the water? How many seconds?

Answer 1

Set the given equation equal to zero (0), since that's the height of the rock when it hits the water:

0 = -16(x^2 - 3x - 18)  (Note:  Please write x^2 to denote "square of x.")

Solving this equation for x, we get:

0 = -16(x-6)(x+3), and x=-3 or x=6.  Since x represents time, discard the negative value x=-3.

The rock hits the water after 6 seconds.

Answer 2

It hit the water at 6 seconds (APEX)