A ball is thrown. The height of the ball "h", in feet, after "t" seconds is given by the equation: h=-16t^-20t+240. How long after the ball is thrown will it strike the ground?

Question

Please explain step by step and use quadratic formula to find the answer.

Ps. The first number of the given equation is -16t to the power of 2 (-16t^2-20t+240)

Answer 1

Set h=0 for ground height, then solve for t

-16t^2 -20t +240 = 0

divide by -4

4t^2 +5t - 60 = 0

use the quadratic formula

t= -5/8 + or - (1/8)square root of (25 +4(4)(60)). Ignore the negative square root

=-5/8 + (1/8)sqr(960+25)

= -5/8 + (1/8)sqr(985)

=-5/8 + (1/8)sqr(5(197))= slightly less than -5/8 +(10/8)(sqr10) = about -5/8 +32/8 = 27/8 = 3.3 seconds

the ball's initial velocity is negative, meaning the ball was thrown downward. Then add the effect of gravity, and it should hit the ground in a short time, even when thrown from an initial height of 240 feet.

plug t=2 into the original equation for height. You should get slightly above ground, near zero

-16(2)^2 -20(2) +240 = -64 -40+240 = 136

try t=3

-16(9) -20(3) +240 = -144 -60 +240 = 36

try t=4

-16(16) -20(4) +240 =-256 -80 +240 =-96

the time to hit ground should be between t=3 and 4, but closer to t=3, maybe about 3 1/3 seconds

Answer 2

The height of a pall is modeled by

h = -16t² - 20t + 240

where h is the height in feet and t is the seconds after release.

When will the ball hit ground? An equivalent question is:

What t value will make h = 0?

Or what is the solution to:

0 = -16t² - 20t + 240

We can solve this using the quadratic formula by recognizing that a=-16, b=-20, and c = 240

Plug these values into the quadratic formula and simplify the radicals and fractions to reach two t values.

The one that makes sense given the constraints of the word problem is the correct answer.