The height of a golf ball that is hit into the air with an initial vertical velocity of 112 feet per second is given by the equation h=-16t^2 + 112t, where t is the time in seconds. After how many seconds is the ball at a height of 96 feet above the ground?
To find the time at which the ball is at a height of 96 feet above the ground, we need to solve the equation h = 96.
Given the equation h = -16t^2 + 112t, we substitute 96 for h:
96 = -16t^2 + 112t
To solve for t, we rearrange the equation to form a quadratic equation in the form of ax^2 + bx + c = 0:
-16t^2 + 112t - 96 = 0
Dividing the entire equation by -16 to simplify:
t^2 - 7t + 6 = 0
Now we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = -7, and c = 6. To solve this equation, we can use factoring or the quadratic formula.
Factoring the quadratic equation:
(t - 1)(t - 6) = 0
By setting each factor equal to zero, we find two possible solutions for t:
t - 1 = 0 --> t = 1
t - 6 = 0 --> t = 6
Hence, the ball is at a height of 96 feet above the ground after 1 second and after 6 seconds.