ball is thrown upwards at 20 ft/sec, the height h after t seconds is given by h = -16t^2 + 20t. How many seconds does it take for the ball to reach its maximum height, and what is this height?
The vertex of this downwards-opening parabola lies halfway between the t-intercepts
these can be found quite easily:
h = -4t(4t - 5)
so t = 0 or t = 5/4
so halfway would be t = 5/8
when t=5/8, h = -16(25/64) + 20(5/8) = 6.25
so the max height of 6.25 ft will be reached after 5/8 seconds
To find the maximum height reached by the ball and the time it takes to reach that height, we need to determine the vertex of the parabolic equation h = -16t^2 + 20t.
The vertex of a parabola in the form h = at^2 + bt + c can be found using the formula:
t = -b / (2a)
In this case, a = -16 and b = 20. Plugging in these values into the formula, we get:
t = -20 / (2 * -16)
t = -20 / -32
t = 0.625 seconds
So, it takes approximately 0.625 seconds for the ball to reach its maximum height.
To find the height at the maximum point, substitute the value of t we just found into the equation:
h = -16t^2 + 20t
h = -16(0.625)^2 + 20(0.625)
h = -16(0.390625) + 12.5
h = -6.25 + 12.5
h = 6.25 feet
Therefore, the ball reaches its maximum height of 6.25 feet after approximately 0.625 seconds.