# Physics

An arrow is shot from a height of 1.4 m toward a cliff of height H. It is shot with a velocity of 28 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.6 s later.

a) What is the height of the cliff (in m)?
I got 25.191 m and that's the correct answer.

(b) What is the maximum height (in m) reached by the arrow along its trajectory?

(c) What is the arrow's impact speed (in m/s) just before hitting the cliff?

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3. 👁 3,758
1. would the max height be 33.9m?

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2. Vo = 28m/s[60o].
Xo = 28*Cos60 = 14 m/s.
Yo = 28*sin60 = = 24.2 m/s.

b. Y^2 = Yo^2 + 2g*h.
0 = 24.2^2 - 19.6h, h = 29.9 m.

c. Y^2 = Yo^2 + 2g*h = 24.2^2 + 19.6(29.9-25.19) = 678, Y = 26.0 m/s = Vertical component.

V = Sqrt(Xo^2+Y^2) =

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