An arrow is shot from a height of 1.5 m toward a cliff of height . It is shot with a velocity of 30 m/s at an angle of 60 deg 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?
you know the height y is
y(t) = 1.5 + 30*.866t - 4.9t^2
= 1.5 + 25.98t - 4.9t^2
y(4) = 27.02
max height is y(2.65) = 35.94
vertical speed is 25.98-9.8*4 = -13.22
horizontal speed is 15.00
So, speed at impact is 20.00 m/s
this sucks no help
To solve this problem, we can use the equations of projectile motion. Let's break down each part of the problem:
(a) What is the height of the cliff?
To find the height of the cliff, we need to consider the vertical motion of the arrow. The arrow is shot from a height of 1.5 m and lands on the top edge of the cliff 4.0 s later. We can use the formula:
Δy = voy * t + (1/2) * a * t^2
where Δy is the change in height, voy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (which is -9.8 m/s^2).
In this case, Δy is the height of the cliff, voy is the vertical component of the initial velocity, t is 4.0 s, and a is -9.8 m/s^2.
We know that the initial velocity has a vertical component equal to v0 * sin(θ), where v0 is the initial velocity (30 m/s) and θ is the angle above the horizontal (60 degrees). We can now substitute the values into the equation to find the height of the cliff as follows:
Δy = (30 m/s * sin(60 deg)) * (4.0 s) + (1/2) * (-9.8 m/s^2) * (4.0 s)^2
Simplifying:
Δy = 52 m
Therefore, the height of the cliff is 52 meters.
(b) What is the maximum height reached by the arrow along its trajectory?
To find the maximum height reached, we need to consider the vertical motion of the arrow. At the highest point, the vertical velocity of the arrow is zero. We can use the equation:
voy = v0y + a * t
where voy is the final vertical velocity, v0y is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
We can solve for t by setting voy to zero and substituting the values:
0 = (30 m/s * sin(60 deg)) + (-9.8 m/s^2) * t
Solving for t:
t = (30 m/s * sin(60 deg)) / 9.8 m/s^2
t ≈ 3.06 s
Now that we have the time it takes to reach the maximum height, we can use the equation for vertical displacement to find the maximum height:
Δy = voy * t + (1/2) * a * t^2
Substituting the values:
Δy = (30 m/s * sin(60 deg)) * (3.06 s) + (1/2) * (-9.8 m/s^2) * (3.06 s)^2
Simplifying:
Δy ≈ 22.8 m
Therefore, the maximum height reached by the arrow along its trajectory is approximately 22.8 meters.
(c) What is the arrow’s impact speed just before hitting the cliff?
To find the arrow's impact speed, we can use the horizontal motion of the arrow. The arrow travels horizontally for a time of 4.0 s, so we can use the equation:
Δx = vox * t
where Δx is the horizontal displacement, vox is the horizontal component of the initial velocity, and t is the time.
We know that the initial velocity has a horizontal component equal to v0 * cos(θ), where v0 is the initial velocity (30 m/s) and θ is the angle above the horizontal (60 degrees). We can now substitute the values into the equation to find the horizontal displacement as follows:
Δx = (30 m/s * cos(60 deg)) * (4.0 s)
Simplifying:
Δx ≈ 60 m
Now that we have the horizontal displacement, we can determine the impact speed by dividing the horizontal displacement by the time it took to cover that distance:
Speed = Δx / t
Speed = 60 m / 4.0 s
Speed = 15 m/s
Therefore, the arrow's impact speed just before hitting the cliff is 15 meters per second.