SOLUTION: A ball is kicked off of the roof of a 40 foot tall building with an initial vertical velocity of 16 feet per second . The balls distance from the ground can
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Question 1081630: A ball is kicked off of the roof of a 40 foot tall building with an initial vertical velocity of 16 feet per second . The balls distance from the ground can be modeled by the equation h =- 16t^2+ vt+s.
A. Find the height of the ball after 1.5 seconds .
B. Determine the maximum height of the ball and the time that passes before the ball reaches that height
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A ball is kicked off of the roof of a 40 foot tall building with an initial vertical velocity of 16 feet per second . The balls distance from the ground can be modeled by the equation h =- 16t^2+ vt+s.
vt = 16; s = 40
h = -16t^2 + 16t + 40
A. Find the height of the ball after 1.5 seconds .
t = 1.5
h = -16(1.5^2) + 16(1.5) + 40
h = -16(2.25) + 24 + 40
h = -36 + 64
h = 28 ft after 1.5 sec
B. Determine the maximum height of the ball and the time that passes before the ball reaches that height
Max occurs on the axis of symmetry. find it; x = -b/(2a)
t = .5 seconds
Find the height at .5 sec
h = -16(.5^2) + 16(.5) + 40
h = -16(.25) + 48
h = -4 + 48
h = 44 ft is the max height
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