Question

A ball is thrown in the air from a ledge. Its height in feet is represented by f(x) = –16(x2 – 7x – 8), 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

$f(x)=-16( x^{2} -7x-8)$

is the functions that determines the height of the ball in Feet, after x seconds being thrown.

so for example to calculate how many feet above the ground in the ball 5 seconds after being thrown, we calculate f(5)

$f(5)=-16( 5^{2} -7*5-8)=-16(25-35-8)=-16(-18)=288$ feet

to solve our problem we need to find x such that f(x)=0

so we solve :

$-16( x^{2} -7x-8)=0$

$x^{2} -7x-8=0$

using the quadratic formula, le a=1, b=-7, c=-8

$\sqrt{D} = \sqrt{ b^{2}-4ac}= \sqrt{ (-7)^{2}-4(1)(-8) }= \sqrt{49+32}= \sqrt{81}=9$

so 

$x_1= \frac{-b+ \sqrt{D} }{2a}= \frac{7+9}{2}= \frac{16}{2}=8$

$x_1= \frac{-b- \sqrt{D} }{2a}= \frac{7-9}{2}= \frac{-2}{2}=-1$, which cannot be the solution to our problem.


Answer: 8 s

Answer 2

You are right thanks