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. After how many seconds will the ball be 52 ft above the ground?

Answer 1

To find the time at which the ball will be 52 ft above the ground, we solve the equation h = -16t^2 + 64t = 52 using the quadratic formula.

Step-by-step explanation:

To find the time at which the ball will be 52 ft above the ground, we need to solve the equation h = -16t^2 + 64t = 52. Rearranging the equation, we get -16t^2 + 64t - 52 = 0. Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / 2a, we can substitute the values a = -16, b = 64, and c = -52 into the formula to find the solutions for t. After calculating, we find that the ball will be 52 ft above the ground at approximately t = 3.54 seconds and t = 0.21 seconds.