I got this problem I can't wrap my head around. help?
H(T) =-4.9t^t + 40t
when hight (H) is 75, what is seconds (T)
I think this is what the equation is:
H(t) = -4.9t^2 + 40t
We want to know what t is when h(t) = 75
So plug in 75 for h(t) and solve for t.
75 = -4.9t^2 + 40t
Subtract 75 from both sides.
0 = -4.9t2 + 40t - 75
Now we can use the quadratic formula to solve for t.
\(t = {-40 \pm \sqrt{40^2-4(-4.9)(-75)} \over 2(-4.9)}\\~\\t={-40 \pm \sqrt{130} \over -9.8}\\~\\ t=\frac{-40+\sqrt{130}}{-9.8}\qquad\text{or}\qquad t=\frac{-40-\sqrt{130}}{-9.8}\)
t ≈ 5.245 seconds or t ≈ 2.918 seconds
Maybe this graph can help you wrap your head around it
Imagine you threw a baIl up in the air and every second you wrote this information down: (what second it was, the height of the ball at that second). For example....at second 1, the height was 35.1 units...and you would write down: (1, 35.1).
Then you plotted all these points on a graph and drew a red line connecting the points. This is what your line would look like:
And the question is, " After how many seconds was the ball was 75 units up in the air? "