I need help with a word problem

Question

An object is launched from a window ledge on a tall building. The height of the object above the ground can be found using the function h(t) = -49/10 t² + 196/10 t + 588/10 where t stands for the number of seconds and h is the height in meters

Question: Solve the equation you made to find the time it takes the object to reach the ground

Answer 1

h(t)=(-49/10)t^2+(196/10)t+588/10

when h(t)=0 the object reaches the ground

factor the equation and set the factors equal to 0

49=7*7 and 588=2*2*3*7*7

2*2*3*7*7*7*7 must be grouped so that the 2 factors subtract to 196/10

first just work with the 196

after some trial and error we get...

(2*3*7*7)*(2*7*7)=294*98 and 294-98=196

(-7t )(7t )=0 so far

the 2 end numbers must multiply to get 588

294/7=42 and 98/7=14 and 42*14=588

(-7t 42)(7t 14)=0 so far

now put in the + and - signs and the 2s and 5s to get the denominators of 10

[(-7t/2)+(42/2)][(7t/5)+(14/5)]=0

(-7t/2)+(42/2)=0

-7t/2=-42/2

multiply both sides by -2

7t=42

t=6 seconds is the answer

Answer 2

You want to set this equation equal to zero because when the height = 0, then the object has hit the ground. I would graph the function h(t) = -49/10 t^2 + 196/10 t + 588/10 and find where it crosses the x-axis.

When you graph it, you can see that it crosses the x-axis at -2 and 6. -2 doesn't make sense for a length of time, so the object hits the ground at t = 6 seconds