To solve the problem, you have correctly set up the equation:
80 = -5t^2 + 20t + 100
To finish solving the equation, you can follow these steps:
1. Simplify the equation:
0 = 5t^2 - 20t - 20
2. Divide all terms by 5 to reduce it:
0 = t^2 - 4t - 4
3. Since this equation does not factor easily, you can use the quadratic formula:
t = (-b Β± β(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -4, and c = -4. Plugging these values into the formula, you get:
t = (-(-4) Β± β((-4)^2 - 4(1)(-4))) / (2(1))
t = (4 Β± β(16 + 16)) / 2
t = (4 Β± β32) / 2
t = (4 Β± 4β2) / 2
t = 2 Β± 2β2
Simplify:
t1 = 2 + 2β2
t2 = 2 - 2β2
These are the two possible solutions for t. You can check their validity by plugging them back into the original equation.
Remember that when t = 0, h = 100, so the positive value of t, t1, represents the first time the ball reaches the desired height.
Therefore, the ball will reach a height of 80 m after approximately 2 + 2β2 seconds (t1) from the time it was thrown.