a ball is thrown upward with an initial velocity of 96 ft/sec from a height 640ft. its height s, in feet, after t seconds is given by s=-16t2+96t+640

Question

I do not understand how this question ends up as 12+- the square root of 784-s all over 4

Answer 1

As Mark stated, you need to tell us what you are trying to determine.

If you are trying to determine how much time it will take for the ball to hit the ground, Mark

is correct in that it will take 10 seconds.  We can discard the -4 seconds.  However, as this

equation is factorable there is an easier way to work it out than using the quadratic equation.

 

First, realize all coefficients are divisible by 16 so factor out the 16.  Actually, to make the squared

term positive, factor out -16.

 

-16t²+96t+640 = -16(t²-6t-40)

 

Setting that equal to zero we have:

 

-16(t²-6t-40)=0

divide both sides by -16 giving

t²-6t-40=0

 

We can factor this quadratic also.. the factors of -40 that add to -6 are -10,4 so...

(t-10)(t+4) = 0

This is only true if either t-10 = 0  or   t+4 = 0 so t = 10 or t=-4

 

As stated, we can disregard t=-4.  The reason we got a negative answer is because

the ball actually only hits the ground one time at the end of its trajectory since it

started 640 ft above ground on an initially upward trajectory.  Thus t=10 seconds

 

Answer 2

Assuming you are to determine the zeros, when it shall return to the earth.

I think the quadratic formula has been used. 

 

t = (−b ± √(b² - 4ac) ) / 2a

 

0 = -16t² + 96t + 640                  Factor out the GCF, -16

0 = -16(t² - 6x - 40)

0 = t² - 6x - 40                         a = 1, b = -6, c = -40

 

t = (6 ± √((-6)² - (4)(1)(-40))) / 2(1)

t = (6 ± √(36 + 160)) / 2

t = (6 ± √196) / 2

t = (6 ± 14) / 2

 

{-4, 10}

 

If some other values is to be determined, please provide additional information.