Question

The height h, in feet, of a ball above the ground t seconds after being thrown upward with a velocity of 64 ft/s is given by h = −16t2 64t 4. After how many seconds will the ball be 52 ft above the ground?

a) 1.5, 2.5, 3.5, 4.5

b) 2, 3, 4, 5

c) 1, 2, 3, 4

d) 0.5, 1.5, 2.5, 3.5

Answer 1

To find when the ball is 52 ft above the ground, the quadratic equation \(-16t^2 + 64t + 4 = 52\) is simplified and factored to yield \(t = 3\) seconds and \(t = 1\) second as the times the ball reaches that height.

Step-by-step explanation:

To solve for the time t when the ball is 52 ft above the ground, we can use the equation h = -16t2 + 64t + 4. We will substitute h with 52 ft and solve for t using the quadratic formula.

The equation becomes 52 = -16t2 + 64t + 4, which simplifies to -16t2 + 64t - 48 = 0. Dividing all terms by -16 to simplify, we get t2 - 4t + 3 = 0.

Factoring the quadratic, we get (t - 3)(t - 1) = 0, which gives us the solutions t = 3 seconds and t = 1 second. Therefore, the ball will be 52 ft above the ground at 1 second and at 3 seconds after being thrown upward.