A kickball is kicked upward with an initial vertical velocity of 3.2 meters
per second. The height of the ball is given by
h=-9.8t^2 + 3.2t
where the height h is measured in feet, and the time t is measured in seconds.
After how many seconds does the ball land?
The height will be 0. Insert into the equation and solve for t.
To find the time at which the ball lands, we need to find the value of t when the height h is equal to 0.
Given the equation h = -9.8t^2 + 3.2t, we can set h = 0 and solve for t:
0 = -9.8t^2 + 3.2t
Now, we have a quadratic equation. To solve it, we can either factor it or use the quadratic formula. In this case, we'll use the quadratic formula since the equation can't be easily factored:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -9.8, b = 3.2, and c = 0. Plugging these values into the formula, we get:
t = (-3.2 ± √(3.2^2 - 4(-9.8)(0))) / (2(-9.8))
Simplifying further:
t = (-3.2 ± √(10.24)) / (-19.6)
t = (-3.2 ± 3.2) / (-19.6)
Now we have two possible values for t:
t = (-3.2 + 3.2) / (-19.6) -> t = 0 / (-19.6) -> t = 0
and
t = (-3.2 - 3.2) / (-19.6) -> t = -6.4 / (-19.6) -> t = 0.3265 (approximately)
Since time cannot be negative in this context, we can discard the negative value. Therefore, the ball lands after approximately 0.3265 seconds.