
#1
Sep2509, 11:24 AM

P: 6

A motorcycle daredevil wants to set a record for jumping over burning school buses. He has hired you to help with the design. He intends to ride off a horizontal platform at 40m/s, cross the burning buses in a pit below him, then land on a ramp sloping down at 20degrees. It's very important that he not bounce when he hits the landing ramp because that could cause him to lose control and crash. You immediately recognize that he won't bounce if his velocity is parallel to the ramp as he touches down. This can be accomplished if the ramp is tangent to his trajectory and if he lands right on the front edge of the ramp. There's no room for error! Your task is to determine where to place the landing ramp. That is, how far from the edge of the launching platform should the front edge of the landing ramp be horizontally and how far below it? There's a clause in your contract that requires you to test your design before the hero goes on national television to set the record.
1. The problem statement, all variables and given/known data V[tex]_{xo}[/tex] = 40m/s V[tex]_{yo}[/tex] = 0m/s 2. Relevant equations I understand that I must find the point on the trajectory where the line tangent to the point is 20[tex]^{o}[/tex]. But how exactly do I figure that out? 3. The attempt at a solution I understand what it takes to solve this problem, my issue is how I figure out which point on the trajectory where the line tangent to the point is 20[tex]^{o}[/tex]. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Sep2509, 11:27 AM

P: 595

The angle depends on the x and y components of the velocity at that point. Assuming no air resistance, the xcomponent is fixed, while the ycomponent can be determined in terms of other variables such as time using the standard kinematics equations.




#3
Sep2509, 11:36 AM

P: 6

The velocity on the xdirection is always constant, so I have to find the point where the velocity in the ydirection is 40tan20? err 40tan20...




#4
Sep2509, 11:45 AM

P: 6

A "Different" kind of kinematics problem..
I guess it is...wow this was simpler than i thought haha.
That's all I needed to know. Thanks! 


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