if a stone is tossed from the top of a 330 meter building, the height of the stone as a function of time is given by h(t)=-0.8t^2-10t+330, where t is in seconds, and height is in meters. After how many seconds will the stone hit the ground?
You must find where is :
-0.8t^2-10t+330=
In google type:
"quadratic equation online"
When you see list of results click on:
webgraphingcom/quadraticequation_quadraticformula.jsp
When page be open in rectacangle type:
-0.8t^2-10t+330=0
and click option solve it.
Solutions are t=(-55/2) and t=15
Time can't be negative so t=15
Proof:
h(t)=h(15)=-0.8*(15^2)-10*15+330=
-0.8*225-150+330=
-180-150+330=-330+330=0
You must find where is :
-0.8t^2-10t+330=0
319.36
To find the time at which the stone hits the ground, we need to find the value of t for which the height h(t) is equal to zero.
Given the equation for the height of the stone as a function of time:
h(t) = -0.8t^2 - 10t + 330
We can set h(t) to zero and solve for t:
0 = -0.8t^2 - 10t + 330
This equation is a quadratic equation in the form of at^2 + bt + c = 0, where a = -0.8, b = -10, and c = 330.
To solve this equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
t = (-(-10) ± sqrt((-10)^2 - 4(-0.8)(330))) / (2(-0.8))
Simplifying further:
t = (10 ± sqrt(100 + 1056)) / (-1.6)
t = (10 ± sqrt(1156)) / (-1.6)
t = (10 ± 34) / (-1.6)
Now we have two possible solutions for t:
t1 = (10 + 34) / (-1.6) ≈ -27.5
t2 = (10 - 34) / (-1.6) ≈ 22.5
Since time cannot be negative in this context, the stone will hit the ground after approximately 22.5 seconds.