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December 22, 2014

December 22, 2014

Posted by **olivia** on Monday, October 1, 2012 at 6:40pm.

a. In how many seconds does it attain maximum height?

b. What is the maximum height?

c. What is the speed of the debris as it reaches 12 ft on the way back to the ground?

d. What is the velocity of the debris at 20 ft on the way up

- calculus and physics -
**Damon**, Monday, October 1, 2012 at 6:57pmEasy way is to assume that total energy is constant (of course debris might have a lot of air drag but there is not much we can do about that.

mass m (will cancel everywhere in this problem)

(1/2) m v^2 + m g h = total energy

at start h = 0 and v = 75

so

(1/2)(75)^2 = total energy/m = 2813

in obsolete English units as here g = 32 ft/s^2

so at 12 ft

(1/2) v^2 = 32(12)

solve for v

and at 20 ft

(1/2) v^2 = 32 (20) solve for v

note that the speed is the same upwards and downwards at the same height, just the direction changes.

- Sorry, went too fast -
**Damon**, Monday, October 1, 2012 at 7:00pmat top

(1/2) v^2 = 0

so

32 h = 2813

at 12 feet

(1/2)v^2 + 32 (12) = 2813

at 20 feet

(1/2) v^2 + 32 (20) = 2813

- calculus and physics -
**Reiny**, Monday, October 1, 2012 at 7:37pma)

Height(t) = -16t^2 + 75t

Height ' (t) = -32t + 75 = 0 for a max of Height

-32t = -75

t = 75/32 seconds or appr 2.34 seconds

b) Height(75/32) = -16(75/32)^2 + 75(75/32) = 87.89 ft

c) when Height = 12

12 = -16t^2+ 75t

16t^2 - 75t + 12 = 0

t = 4.5216 sec (on the way down) or

t = .1659 seconds (on the way up)

velocity(t) = height ' (t)

= -32t + 75 , so we want it on the way down

velocity(4.5216) = -32(4.5216)+75 = -69.69 ft/sec

d)

same method as in c)

16t^2 - 75t + 20 = 0

t = 4.40 seconds (on the way down), or

t = .284 secpmds (on the way up

velocity(.284) = -32(.284)+75 = 65.92 ft/sec

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