To solve this problem, we can break it down into three steps:
Step A: Find the magnitude of Eddie's initial velocity.
Step B: Determine Eddie's initial direction of motion relative to the horizontal.
Step C: Calculate the height of the ramp's edge relative to where Eddie landed.
Let's start with Step A:
Step A: Find the magnitude of Eddie's initial velocity.
We can use the horizontal displacement and velocity just before landing to find the time of flight. Then, we can use this time to find the magnitude of the initial velocity.
Given:
Horizontal displacement, Δx = 53.0 m
Velocity just before landing, v = 26.0 m/s
The horizontal displacement can be determined using the following equation:
Δx = v_x * t, where v_x is the horizontal component of the velocity and t is the time of flight.
Since the horizontal component of the velocity does not change during flight, we can write:
v_x = v * cos(37.0°), where v is the magnitude of the velocity.
Rearranging the equation, we can solve for t:
t = Δx / (v * cos(37.0°))
Substituting the given values:
t = 53.0 m / (26.0 m/s * cos(37.0°))
Using a calculator, we find
t ≈ 1.59 s
Now, let's find the magnitude of the initial velocity, |v0|:
Using the equation of motion:
Δx = v0 * cos(θ) * t,
where θ is the angle of the velocity vector with the horizontal and t is the time of flight.
Substituting the given values and solving for |v0|:
|v0| = Δx / (cos(θ) * t)
= 53.0 m / (cos(37.0°) * 1.59 s)
Using a calculator, we find
|v0| ≈ 41.34 m/s
So, the magnitude of Eddie's initial velocity as he left the ramp is approximately 41.34 m/s.
Now, let's move on to Step B:
Step B: Determine Eddie's initial direction of motion relative to the horizontal.
We can use the given angle below the horizontal and the angle made by the velocity vector with the horizontal to find the initial direction of motion relative to the horizontal.
Given:
Angle below the horizontal, θ1 = 37.0°
The angle made by the velocity vector with the horizontal can be found using the equation:
θ = tan^(-1)(v_y / v_x), where v_y is the vertical component of the velocity and v_x is the horizontal component of the velocity.
In this case, since Eddie's velocity is pointing below the horizontal, we can write:
θ = tan^(-1)(-v_y / v_x)
Substituting the given values, we have:
θ = tan^(-1)(-v * sin(37.0°) / v * cos(37.0°))
Simplifying, we find:
θ ≈ tan^(-1)(-sin(37.0°) / cos(37.0°))
Using a calculator, we find:
θ ≈ -53.0°
Therefore, Eddie's initial direction of motion as he left the ramp, measured relative to the horizontal, is approximately -53.0°.
Finally, let's proceed to Step C:
Step C: Calculate the height of the ramp's edge relative to where Eddie landed.
Since we know the horizontal and vertical displacements, we can use the equation of motion to find the initial vertical velocity and then calculate the height of the ramp's edge.
Using the equation of motion in the vertical direction:
Δy = v0 * sin(θ) * t - (1/2) * g * t^2,
where Δy is the vertical displacement, v0 is the magnitude of the initial velocity, θ is the angle above the horizontal, t is the time of flight, and g is the acceleration due to gravity.
Given:
Vertical displacement, Δy = 0 m (since Eddie lands at the same height as the ramp's edge)
Angle above the horizontal, θ2 = 90° - θ1 (since θ1 is measured below the horizontal)
Acceleration due to gravity, g ≈ 9.8 m/s^2
Substituting the given values, we have:
0 = |v0| * sin(90° - 53.0°) * 1.59 s - (1/2) * 9.8 m/s^2 * (1.59 s)^2
Simplifying, we find:
0 = |v0| * sin(37.0°) * 1.59 s - 9.8 m/s^2 * (1.59 s)^2
Solving for |v0|, we have:
|v0| * sin(37.0°) * 1.59 s = 9.8 m/s^2 * (1.59 s)^2
Using a calculator, we find:
|v0| ≈ 14.58 m/s
Since the height of the ramp's edge is given by:
Height = |v0| * sin(θ2) * t
= 14.58 m/s * sin(37.0°) * 1.59 s
Using a calculator, we find:
Height ≈ 11.67 m
Therefore, the height of the ramp's edge relative to where Eddie landed is approximately 11.67 m.