Saturday

October 25, 2014

October 25, 2014

Posted by **Brenda** on Sunday, June 23, 2013 at 6:15pm.

straight up.

(a) By how much does the speed of the ball decrease

each second while it is ascending?

(b) By how much does its speed increase each second

while it is descending?

(c) How does the time of ascent compare with the time

of descent?

- conceptual science -
**Henry**, Tuesday, June 25, 2013 at 5:32pma. V = Vo-g*t

V = Vo-9.8m/s^2*1s

V = Vo-9.8m/s. s.

The speed decreases by 9.8 m each second.

b. V = Vo + g*t

V = Vo + 9.8m/s^2*1s

V = Vo + 9.8m/s.

The speed increases by 9.8 m each second.

c. V = Vo + g*t = 0 @ max. ht.

Vo + g*t = 0

Vo - 9.8t = 0

9.8t = Vo

Tr = Vo/9.8 = Rise time=Time of ascent.

V = Vo + g*t

V = 0 + 9.8*t

Tf = V/9.8

V = Vo during ascent.

Therefore, Tf = Vo/9.8 = Tr.

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