A skateboarder shoots off a ramp with a velocity of 6.6 m/s,directed at an angle
ID: 1681073 • Letter: A
Question
A skateboarder shoots off a ramp with a velocity of 6.6 m/s,directed at an angle of 58 degrees above horizantal. The end of theramp is 1.2 m above the ground. Let the x axis be parallel to theground, the +y direction be vertically upward, and take as theorigin the point on the ground directly below the top of theramp. a) How high above the ground is the highest point that theskateboarder reaches? b) When the skateboarder reaches the highest point, how far isthis point horizontally form the end of the ramp?For part a, I found the point at which the skateboarder isback at the top of the ramp height by finding when his velocity isequal to -V0y, then dividing this time by two andfinding his height above the ramp and finally adding 1.2 m tothis height to account for the ramp. Is this a legitimate way tosolve this problem? I got 2.8 m for part (a) and 0.403 m for part(b). a) How high above the ground is the highest point that theskateboarder reaches? b) When the skateboarder reaches the highest point, how far isthis point horizontally form the end of the ramp?
For part a, I found the point at which the skateboarder isback at the top of the ramp height by finding when his velocity isequal to -V0y, then dividing this time by two andfinding his height above the ramp and finally adding 1.2 m tothis height to account for the ramp. Is this a legitimate way tosolve this problem? I got 2.8 m for part (a) and 0.403 m for part(b).
Explanation / Answer
vertical acceleration=g=9.8let the skateboarder reach a height h above the end of theramp u=6.6*sin58 v=0
v2=u2-2*g*h 0=(6.6*sin58)2-2*9.8*h h=1.60 m
the highest point is 1.2+1.60=2.8 m above the ground
time taken to reach the highest point=t v=u-g*t 0=6.6*sin58-9.8*t t=0.5711 s
initial horizontal velocity=6.6*cos58 horizontal acceleration=0 distance traveled in 0.5711 shorizontally=6.6*cos58*0.5711=2.00
the highest point is 2.00 m from the end of the ramp