1. A ball is thrown straight up and passes point at a height of 40 m...

Question

Question

1. A ball is thrown straight up and passes point at a height of 40 m above its starting point in 5 s. What was its initial speed?

Answer 1

Answer #1

Kinematic equation

y=vi*t-(.5)gt^2

40 = vi(5) - (0.5)(9.8)(5)^2

40 = vi(5) - 122.5

vi(5) = 162.5

vi = 32.5 m/s

Answer 2

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