A skateboarder shoots off a ramp with a velocity of 5.7 m/s, directed at an angle of 58° above the horizontal. The end of the ramp is 1.3 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
u = horizontal velocity = 5.7 cos 58
= 3.02 m/s forever
Vi = initial speed up = 5.7 sin 58
= 4.83 m/s
v = Vi - 9.8 t
at top, v = o
0 = 4.83 - 9.8 t
t = .493 s at top
h = 1.3 + Vi t - 4.9 t^2
t at top is .493
so
h at top = 1.3 + 4.83(.493)-4.9(.493)^2
= 2.49 meters above ground
x = u t
x = 3.02 * .493 = 1.49 meters
To answer these questions, we can use the principles of projectile motion. We'll break down the initial velocity into its horizontal and vertical components to solve for the maximum height reached and the horizontal distance traveled.
(a) To find the highest point reached by the skateboarder, we need to determine the vertical component of the initial velocity.
The initial velocity, V₀ = 5.7 m/s, is directed at an angle of 58° above the horizontal.
Vertical component of the initial velocity, V₀y = V₀ * sin(θ),
where θ is the angle above the horizontal.
V₀y = 5.7 m/s * sin(58°).
Now, we need to find the time it takes for the skateboarder to reach the highest point of the trajectory. At this point, the vertical component of the velocity becomes zero.
The vertical motion can be described by the equation:
Δy = V₀y * t + (1/2) * g * t²,
where Δy is the change in height, t is the time taken, and g is the acceleration due to gravity (-9.8 m/s²).
We can set the vertical displacement, Δy, equal to zero since the skateboarder reaches the highest point:
0 = V₀y * t + (1/2) * g * t².
Rearranging the equation, we get:
(1/2) * g * t² = -V₀y * t.
Since t is non-zero (time cannot be zero), we can cancel out t and write:
(1/2) * g * t = -V₀y.
Now, we can solve for t:
t = -(2 * V₀y) / g.
Substituting the values:
t = -(2 * 5.7 m/s * sin(58°)) / -9.8 m/s².
Calculate t to get the time taken to reach the highest point.
Once we have the time t, we can substitute it back into the equation for Δy:
Δy = V₀y * t + (1/2) * g * t².
Δy will give us the height above the ground at the highest point.
(b) To find the horizontal distance traveled when the skateboarder reaches the highest point, we can use the horizontal component of the initial velocity.
Horizontal component of the initial velocity, V₀x = V₀ * cos(θ),
where θ is the angle above the horizontal.
V₀x = 5.7 m/s * cos(58°).
The time taken to reach the highest point (from part a) is also the time taken to reach the horizontal distance.
So, we multiply the time by the horizontal component of the velocity to get the horizontal distance traveled.
Horizontal distance traveled = V₀x * t.
Solve this equation to find the horizontal distance.
By following these steps, you can determine the highest point above the ground (a) and the horizontal distance traveled from the ramp to the highest point (b).