A skateboarder shoots off a ramp with a velocity of 6.3 m/s, directed at an angl
ID: 1591987 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 6.3 m/s, directed at an angle of 56° above the horizontal. The end of the ramp is 1.1 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?
Explanation / Answer
Given that
A skateboarder shoots off a ramp with a velocity of (vi) = 6.3 m/s
The ramp directed at an angle of (theta) =56° above the horizontal.
The initial velocity is resolved into two components at anlge of 56 degrees then
Velocity along the x-axis is vx =6.3cos56 =3.523m/s
Velocity along the y-axis is vy =6.3sin56 =5.223m/s
The acceleration due to gravity is (g) =9.81m/s2
a)
When it reached to maximum height final vertical velocity is Vf =0
Therefore
0 =Vf2 =vy2-2gy
Gives the maximum height as
y =vy2/2g =(5.223)2/2*9.81 =1.390m
b)
How far is this point horizontally from the end of the ramp is given by range
Then x =Horizontal component of velocity * time of flight
=vx*t =vx*vy/g =(3.523)(5.223)/(9.81)=1.875m