Posted by **xxxTaraKhanxxx** on Tuesday, September 27, 2011 at 12:14am.

A circus performer is juggling clubs while standing on stilts. He released a club from a point 8 feet above the ground with an initial vertical velocity of 10 ft per second.

a. Write the equation that models the height [h] (in feet) of the club as a function of the time [t] in seconds.

b.How high does the club go? (Round answer to nearest tenth)

c. How long after the club is released does it reach its maximum height? (Round to nearest tenth)

d. When is the club at the height of 9 feet?

e. After the performer releases the club, how much time does he have to catch the club on its way down when the club reaches a height of 7 feet? (Round answers to nearest tenth)

f. If the performer does not catch the club, when will it land on the ground? Round your answer to the nearest tenth.

- Math-Urgent -
**Henry**, Wednesday, September 28, 2011 at 3:58pm
a. h = ho + Vo*t -16t^2.

ho = Initial height.

Vo = Initial velocity.

b. h = ho + (Vf^2 - Vo^2) / 2g,

h = 8 + (0 - (10)^2) / -64.4 = 9.55 Ft.

above ground.

c. t = (Vf - Vo) / g,

t = (0 - 10) / -32.2 = 0.31s.

d. h = 8 + 10t - 16t^2 = 9 Ft.

-16t^2 + 10t + 8 = 9,

-16t^2 + 10t - 1 = 0,

Use Quadratic Formula to find t:

t = 0.125s.

e. d = Vo*t + 0.5gt^2 = 9.55 - 7 = 2.55 Ft.

0 + 16t^2 = 2.55,

t^2 = 0.159,

t(dn) = 0.40s.

T = t(up) + t(dn) = 0.31 + 0.40 = 0.7s to catch the club.

f. d = Vo*t + 16t^2 = 9.55 Ft.

0 + 16t^2 = 9.55,

t^2 =0.597,

t = 0.8s.

- Math-Urgent -
**kala**, Thursday, September 29, 2016 at 11:52am
0.8 seconds

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