Posted by **Ali** on Tuesday, January 24, 2012 at 11:46am.

A stone is launched straight up by a slingshot. Its initial speed is 20.1 m/s and the stone is 1.30 m above the ground when launched. Assume g = 9.80 m/s2.

(a) How high above the ground does the stone rise?

(b) How much time elapses before the stone hits the ground?

- physics -
**Henry**, Thursday, January 26, 2012 at 5:14pm
a. hmax = ho + (Vf^2-Vo^2 / 2g.

hmax=1.3 +(0-(20.1)^2 / -19.6=21.9 m.

b. Tr = Vf-Vo) / g,

Tr = (0-20.1) / -9.8 = 2.05 s. = Rise

time or time to reach max ht.

h = Vo*t + 0.5g*t^2 = 21.9 m.

0 + 4.9t^2 = 21.9,

t^2 = 4.47,

Tf = 2.11 s.Time to fall to ground.

Tr + Tf = 2.05 + 2.11 = 4.16 s. = Time

in flight.

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