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?
2 answers
would the max height be 33.9m?
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) =
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) =
3 answers