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suppose a ball thrown downward with initial velocity 3 m/s instead of being dropped
a)what then would be its position after 1s and 2s ?
b)what would its speed be after 1 s and 2 s ?
13 years ago

Answers

Henry
a. d = Vo*t + 0.5g*t^2,
d=3*1 + 4.9*1^2 = 7.9m below starting point.

d = 3*2 + 4.9*2^2 = 25.6m

b. Vf^2 = V0^2 + 2g*d,
Vf^2 = 3^2 + 19.6*7.9 = 163.84,
Vf = 12.8m/s.

Vf^2 = 3^2 + 19.6*25.6 = 510.76,
Vf = 22.6m.
13 years ago

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