Assume you are throwing a baseball up from the top of a building. The equation for the height of the ball, h, at time, t, is represented by the equation: h=-16t^2+64t+512
(5 points) After how many seconds does the ball reach 560 feet? Explain what your answer means.
h=-16t^2+64t+512
we want h = 560
560 = -16t^2+64t+512
16t^2 - 64t + 48 = 0
t^2 - 4t + 3 = 0
(t-1)(t-3) = 0
it happened after 1 second AND after 3 seconds, once on its way up and then again on its way down
To find the number of seconds it takes for the ball to reach a height of 560 feet, we need to solve the equation h = 560.
The equation provided is h = -16t^2 + 64t + 512. We can solve it by substituting h with 560:
560 = -16t^2 + 64t + 512
Next, we need to rearrange the equation to make it equal to zero:
-16t^2 + 64t + 512 - 560 = 0
Simplifying further:
-16t^2 + 64t - 48 = 0
Now, we have a quadratic equation in terms of t. We can solve it by factoring, completing the square, or using the quadratic formula.
Using factoring:
-16t^2 - 48 + 64t = 0
-16t(t - 4) + 16(t - 4) = 0
(16 - 16t)(t - 4) = 0
Setting each factor equal to zero:
16 - 16t = 0 or t - 4 = 0
Solving each equation:
16 - 16t = 0
16t = 16
t = 1
t - 4 = 0
t = 4
So, the ball reaches a height of 560 feet after 1 second and again after 4 seconds.
What this means is that the ball reaches a height of 560 feet twice during its trajectory. It first reaches that height 1 second after being thrown, and then it reaches the same height again 4 seconds after being thrown.