Question

A ball is thrown in the air from a ledge. Its height in tesspresented by

R -16(x^2
- 5x-6), where x is the number of seconds since the ball has
been thrown. The height of the ball is 0 feet when it hits the ground.
How many seconds does it take the ball to reach the ground?

Answer 1

6 .74 seconds

Explanation:

The height of the ball is given by the equation:

`R(x) = - 16( {x}^(2) - 5x - 6)`

When the ball hit the ground, then;

`R(x) = 0 \\ - 16( {x}^(2) - 5x - 6) = 0 \\ {x}^(2) - 5x - 6 = 0 \\ x = -0.74 \: Or \: 6.74`

We discard x=-0.74 since time is not negative.

Therefore the ball hit the ground after 6.74 seconds.