A)Eddie the Eagle, British Olympic ski jumper, is attempting his most mediocre jump yet. After leaving the end of the ski ramp, he lands downhill at a point that is displaced 53.0 m horizontally from the edge of the ramp. His velocity just before landing is 26.0 m/s and points in a direction 37.0 degree below the horizontal. Neglect any effects due to air resistance or lift.
What was the magnitude of Eddie's initial velocity as he left the ramp?
B)Determine Eddie's initial direction of motion as he left the ramp, measured relative to the horizontal.
C)Calculate the height of the ramp's edge relative to where Eddie landed.
To solve this problem, we will use the principles of projectile motion.
A) Let's start by finding the horizontal component of Eddie's initial velocity as he left the ramp. We can do this using trigonometric functions. The horizontal component is given by:
Vx = V * cos(theta)
where Vx is the horizontal component of the velocity, V is the magnitude of the velocity, and theta is the angle below the horizontal.
Substituting the given values:
V = 26.0 m/s
theta = 37.0 degrees
Vx = 26.0 m/s * cos(37.0 degrees)
Calculating Vx:
Vx = 26.0 m/s * 0.7986
Vx = 20.7644 m/s
B) Next, let's find the vertical component of Eddie's initial velocity as he left the ramp. The vertical component is given by:
Vy = V * sin(theta)
where Vy is the vertical component of the velocity, V is the magnitude of the velocity, and theta is the angle below the horizontal.
Substituting the given values:
V = 26.0 m/s
theta = 37.0 degrees
Vy = 26.0 m/s * sin(37.0 degrees)
Calculating Vy:
Vy = 26.0 m/s * 0.6018
Vy = 15.6428 m/s
C) Now, to find the initial velocity, we can use the Pythagorean theorem. The magnitude of the initial velocity is given by:
V = sqrt(Vx^2 + Vy^2)
Substituting the calculated values:
Vx = 20.7644 m/s
Vy = 15.6428 m/s
V = sqrt((20.7644 m/s)^2 + (15.6428 m/s)^2)
Calculating V:
V = sqrt(431.1858 m^2/s^2 + 244.4936 m^2/s^2)
V = sqrt(675.6794 m^2/s^2)
V ≈ 26.0 m/s
So, the magnitude of Eddie's initial velocity as he left the ramp is approximately 26.0 m/s.
B) To find Eddie's initial direction of motion as he left the ramp relative to the horizontal, we use the inverse tangent function. The initial angle is given by:
theta_initial = arctan(Vy / Vx)
Substituting the calculated values:
Vx = 20.7644 m/s
Vy = 15.6428 m/s
theta_initial = arctan(15.6428 m/s / 20.7644 m/s)
Calculating theta_initial:
theta_initial = arctan(0.7519)
theta_initial ≈ 36.61 degrees
So, Eddie's initial direction of motion as he left the ramp, relative to the horizontal, is approximately 36.61 degrees.
C) Finally, to calculate the height of the ramp's edge relative to where Eddie landed, we can use the equations of projectile motion. First, we need to find the time it took for Eddie to reach the horizontal displacement of 53.0 m.
Using the horizontal component of the velocity,
Vx = 20.7644 m/s
The time of flight (t) is given by:
t = displacement / Vx
Substituting the given values:
displacement = 53.0 m
Vx = 20.7644 m/s
t = 53.0 m / 20.7644 m/s
Calculating t:
t ≈ 2.5525 s
Now, we can use the time of flight to calculate the height.
The height (h) is given by:
h = Vy * t + (1/2) * (-9.8 m/s^2) * t^2
Substituting the given values:
Vy = 15.6428 m/s
t ≈ 2.5525 s
h = 15.6428 m/s * 2.5525 s + (1/2) * (-9.8 m/s^2) * (2.5525 s)^2
Calculating h:
h ≈ 32.0179 m
So, the height of the ramp's edge relative to where Eddie landed is approximately 32.0179 m.