A skateboarder shoots off a ramp with a velocity of 6.9 m/s, directed at an angl
ID: 2046095 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 6.9 m/s, directed at an angle of 55° above the horizontal. The end of the ramp is 1.5 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?
??? m
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
??? m
Explanation / Answer
Part A)
So for this part you are looking for the highest point that the skateboarder reaches. The initial Y value when he leaves the ramp is 1.5m. You can find how much more he goes up by using simple kinematics.
First find the Y component of the velocity when he leaves the ramp.
Let's let V = 6.9m/s
and = 55
Vy = Vsin() (answer should be 5.62m/s)
Now what you need to do is find the time until that Y velocity goes to zero because that's when he reaches the peak of the jump. Acceleration of -9.8m/s2 is always a given.
Use the equation
Vf2 = V02 + 2ay
Here, the final velocity is zero because once again you want the peak. Initial velocity is the velocity you found before, and the acceleration is given. Solve for y.
After you find y, add it to the initial height of 1.5m and you should get your answer.
(answer should be 3.13m)
Now for Part B
This part requires you to find the time it takes from leaving the ramp to when he reaches the peak.
You can do this by using this kinematic equation.
Vf = V0 + at
Use the same values for initial and final y velocities we have above and solve for t (time). Acceleration is the same as well.
After you find this time, you need to find the x component of the velocity which is
Vx = Vcos()
Then just simply multiply this velocity with the time you derived above and you should get the answer.
(answer should be 2.27m)
Hope this helps