In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0...

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In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0...

In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 43.0 ∘ angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground.

How far does the shot travel?

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Answer 1

Answer #1

Vx = 13 cos43 = 9.51 m/s

Vy = 13 sin43 = 8.87 m/s

Time to reach max height, t = 8.87/9.8 = 0.90 sec

H = 8.87 x 0.90 - 0.5 x 9.8 x 0.90^2 = 4.01 m

Maximum height = 4.01 + 1.80 = 5.81 m

Time, to = sqrt(2 x 5.81/9.8) = 1.09 sec

Total time, T = 1.09 + 0.90 = 1.99 sec

Distance, d = 9.51 x 1.99 = 18.91 m

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Answer 2

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In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0...

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