If a ball is thrown straight upwards with a velocity of 80 ft/s, then its height after t seconds is given by s = 80t - 16t2. Find the maximum height in feet reached by the ball
2007-05-26 07:01:25 · 2 answers · asked by chetzel 3 in Science & Mathematics ➔ Mathematics
To get its maximum, take the derivative of "s" and set to zero. You'll get an equation with one unknown t, which is the time after throwing the ball (after t = 0) when maximum height is reached.
s' = 80 - 32t = 0
t = 2.5 seconds
Maximum height is
s = 80*2.5 - 16*(2.5)^2 = 100 feet
2007-05-27 07:49:46 · answer #1 · answered by sweetwater 7 · 0⤊ 0⤋
s = 80t - 16t^2
s = -16(t^2 - 5t)
s = -16(t^2 - 5t + 25/4 - 25/4)
s = -16(t - 5/2)^2 + 200
s - 200 = -16(t - 5/2)^2
This is a parabola oopening down with a vertex at (5/2, 200).
maximum height using this equation is 200 ft.
2007-05-26 07:20:43 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋