The equation y = -16t^2 - 60t + 54$ describes the height (in feet) of a ball thrown downward at 60 feet per second from a height of 54 feet from the ground. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.
+9465 height = -16t2 - 60t + 54 , where t is the number of seconds after the ball is released
In how many seconds will the ball hit the ground?
In other words, what is t when the height is zero?
Plug in 0 for the height and solve for t .
0 = -16t2 - 60t + 54
We can divide both sides of the equation by -2
0 = 8t2 + 30t - 27
Let's split 30t into two terms such that their coefficients multiply to -216
0 = 8t2 - 6t + 36t - 27
Factor 2t out of the first two terms and factor 9 out of the last two terms
0 = 2t(4t - 3) + 9(4t - 3)
Factor (4t - 3) out of both remaining terms
0 = (4t - 3)(2t + 9)
Set each factor equal to 0 and solve for t
| 4t - 3 = 0 | _______or_______ | 2t + 9 = 0 |
|
| 4t = 3 |
| 2t = -9 | |
| t = 3/4 | t = -9/2 |
| |
| t = 0.75 | t = -4.5 |
We want the number of seconds after the ball is released, so we only want the positive solution.
The ball will hit the ground 0.75 seconds.
Here's a graph: https://www.desmos.com/calculator/kznk4jayc4
+9465 Yep! You can use the quadratic formula to solve -16t2 - 60t + 54 = 0 for t , like this:
\(t \,=\, {-(-60) \,\pm \,\sqrt{(-60)^2-4(-16)(54)} \over 2(-16)}\,=\,{60\, \pm \,\sqrt{7056} \over -32}\,=\, {60 \,\pm\, 84 \over -32}\\~\\ \ \\ \begin{array}\ t\,=\,\frac{60+84}{-32}&\qquad\text{or}\qquad& t\,=\,\frac{60-84}{-32}\\~\\ t\,=\,\frac{144}{-32}&\qquad\text{or}\qquad& t\,=\,\frac{-24}{-32}\\~\\ t\,=\,-4.5&\qquad\text{or}\qquad& t\,=\,0.75 \end{array}\)
