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An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow's impact speed just before hitting the cliff?

Respuesta :

ijustinkim

Known Variables:

Δy = 1.5 m

v0 = 30m/s at 60° above the horizontal.

t = 4.0 s

a. vfy = v0y + at

0 = 30sin(60) + (-9.8)t

-25.98 = -9.8t

t = 2.65.

4-2.65 = 1.35s.

Δy = v0yt + at^2/2

Δy = 25.98(2.65) + (-9.8)(2.65)^2/2

Δy = 34.44 m

1.5 + 34.44 = 35.94 m.

Δy = v0t + at^2/2

Δy = (-9.8)(1.35)^2/2

Δy = -8.93.

35.94 - 8.93 = 27.01 m.

b. vfy = v0y + at

0 = 30sin(60) + (-9.8)t

-25.98 = -9.8t

t = 2.65.

Δy = v0yt + at^2/2

Δy = 25.98(2.65) + (-9.8)(2.65)^2/2

Δy = 34.44 m

1.5 + 34.44 = 35.94 m.

vfy = v0y + at

vfy = (-9.8)(1.35)

vfy = -13.23 m/s.

I might have gone wrong somewhere so please check my answer! Cheers!

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