If a ball is thrown vertically upward from the ground, with an initial velocity of 64 ft/s, it's height h after t seconds is given by h=-16t^2+64t how long does it take the ball to return to the ground?
To find out how long it takes for the ball to return to the ground, we need to determine the time when the height (h) becomes zero.
Given that the height of the ball is represented by the equation h = -16t^2 + 64t, we can set h = 0 and solve for t.
0 = -16t^2 + 64t
To solve this quadratic equation, we can factorize it or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For the equation -16t^2 + 64t = 0, the coefficients are:
a = -16
b = 64
c = 0
Plugging these values into the quadratic formula, we get:
t = (-64 ± √(64^2 - 4(-16)(0))) / (2(-16))
Simplifying further:
t = (-64 ± √(4096)) / -32
t = (-64 ± 64) / -32
Now, let's consider both values separately:
For t = (-64 + 64) / -32:
t = 0 / -32
t = 0
For t = (-64 - 64) / -32:
t = -128 / -32
t = 4
Therefore, the ball takes 0 seconds and 4 seconds to return to the ground.