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?
would the max height be 33.9m?
To find the maximum height reached by the arrow along its trajectory, we can use the equations of kinematics.
We know the initial velocity (v₀ = 28 m/s) and the launch angle (θ = 60°). The vertical component of the initial velocity can be found using the equation:
v₀y = v₀ * sin(θ)
v₀y = 28 m/s * sin(60°) = 24.248 m/s
The time taken for the arrow to reach its maximum height can be found using the equation:
t = v₀y / g
Where g is the acceleration due to gravity, approximately 9.8 m/s².
t = 24.248 m/s / 9.8 m/s² = 2.48 s
The maximum height reached can be found using the equation:
h = v₀y * t - 0.5 * g * t²
h = 24.248 m/s * 2.48 s - 0.5 * 9.8 m/s² * (2.48 s)²
h ≈ 29.9 m
Therefore, the maximum height reached by the arrow is approximately 29.9 m.
To find the arrow's impact speed just before hitting the cliff, we can use the horizontal component of the initial velocity and the time taken for the arrow to reach the cliff.
The horizontal component of the initial velocity can be found using the equation:
v₀x = v₀ * cos(θ)
v₀x = 28 m/s * cos(60°) = 14 m/s
The distance traveled horizontally can be found using the equation:
d = v₀x * t
d = 14 m/s * 3.6 s = 50.4 m
The impact speed can be found using the equation:
v = d / t
v = 50.4 m / 3.6 s
v ≈ 14 m/s
Therefore, the arrow's impact speed just before hitting the cliff is approximately 14 m/s.
To solve these questions, we can break down the motion of the arrow into its horizontal and vertical components.
a) To find the height of the cliff (H), we can use the vertical component of the motion. The equation we need to use is the vertical displacement formula:
y = y0 + v0y * t + (1/2) * a * t^2
Here, y is the final height (1.4 m), y0 is the initial height (0 m), v0y is the vertical component of the initial velocity, t is the time taken (3.6 s), and a is the acceleration due to gravity (-9.8 m/s^2).
To find v0y, we use the initial velocity (28 m/s) and the angle of projection (60 degrees):
v0y = v0 * sin(theta)
= 28 m/s * sin(60 degrees)
= 28 * sqrt(3) / 2
= 14 sqrt(3) m/s
Now we can substitute the values into the equation:
1.4 m = 0 m + (14 sqrt(3) m/s) * (3.6 s) + (1/2) * (-9.8 m/s^2) * (3.6 s)^2
Simplifying and solving for H, we get:
H = 1.4 m - 56.7 m + 62.86 m
H = 25.191 m (rounded to three decimal places)
So, the height of the cliff is 25.191 m.
b) To find the maximum height reached by the arrow, we can use the vertical component of the motion. At the maximum height, the vertical velocity component becomes zero.
v0y - gt = 0
Substituting the values:
14 sqrt(3) m/s - 9.8 m/s^2 * t = 0
Solving for t:
t = (14 sqrt(3) m/s) / (9.8 m/s^2)
t ≈ 2.04 s
Now we can use this time to find the maximum height using the vertical displacement formula:
y = y0 + v0y * t + (1/2) * a * t^2
y = 0 m + (14 sqrt(3) m/s) * (2.04 s) + (1/2) * (-9.8 m/s^2) * (2.04 s)^2
Simplifying and calculating, we get:
y ≈ 15.912 m (rounded to three decimal places)
So, the maximum height reached by the arrow is approximately 15.912 m.
c) To find the impact speed of the arrow just before hitting the cliff, we can use the horizontal component of the motion. The impact speed will be the horizontal component of the final velocity.
The horizontal velocity remains constant throughout the motion, so we can use the formula:
v = v0x
where v is the final velocity, and v0x is the horizontal component of the initial velocity.
v0x = v0 * cos(theta)
= 28 m/s * cos(60 degrees)
= 28 m/s * 0.5
= 14 m/s
So, the impact speed of the arrow just before hitting the cliff is 14 m/s.
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) =