(a) vertex of the parabola is at t = -b/2a
(b) plug in that value to get h
(c) solve for t when h=0
A ball is thrown upward from a height of 240 feet. the height h of the object(in feet) t seconds after the ball is released is given by h=-16t^2+32t+240
a. how long does it take the ball to reach its maximum height?
b. what is the maximum height attained by the ball?
c. how long does it take the object to hit the ground?
2 answers
a) x = -b/2a
x = -32/2(-16) = 1
t = 1 sec to reach max height
b) h = -16(1)^2 + 32(1) + 240
h = -16 + 32 + 240 = 256
h = 256 ft is the max height
c) 0 = -16(t^2 -2t-15)
0 = -16(t-5)(t+ 3)
t = -3, t = 5
t = 5 seconds for the ball to hit the ground.
x = -32/2(-16) = 1
t = 1 sec to reach max height
b) h = -16(1)^2 + 32(1) + 240
h = -16 + 32 + 240 = 256
h = 256 ft is the max height
c) 0 = -16(t^2 -2t-15)
0 = -16(t-5)(t+ 3)
t = -3, t = 5
t = 5 seconds for the ball to hit the ground.
2 answers