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- Scott and jane attempt to jump a gap of 135 meters on scott's Harley....(projectile motion)?
1. Scott and jane attempt to jump a gap of 135 meters on scott's Harley. If they leave the jump ramp at a speed of 35m/s and an angle of 45 degrees, do they make the jump? Please indicate which equation you use, what variable corresponds to each number, and how you got the number that goes with the variable. Thanks!

A First, lets look at what we are given and what is required: Horizontally: V1x= 35m/s (cos45) = 24.75m/s V2x= Unknown ax= 0m/s2 (since gravity doesn't pull objects horizontally) dx= Unknown tx= Unknown Vertically: V1y = 35m/s (sin45) = 24.75m/s V2y= Unknown ay= -9.8m/s2 (negative because direction of pull is downwards) dy= -135 m (negative again) ty= Unknown Now, remember that tx = ty, so both times are equal. The formula we are using is d=(v1)(t) + (0.5)(a)(t) First, we solve for t in the y-direction (vertically) dy=(V1y)(t) + (0.5)(ay)(t)2 [squared, not multiplied by two) -135=(24.75)(t) = (0.5)(-9.8)(t)2 Now, since we cannot go further with this equation, we must rearrange it into quadratic form and use the quadratic equation. ax2 + bx + c = 0 0= (-4.9)t2 + 24.75t + 135 t = -b +- (squareroot) b2 - 4ac / 2a (note that it is hard to indicate the square roots and squares, so i trust you are familiar with the quadratic equation, if not look it up and insert the variables) you should get 2 answers since its the quadratic equation: -3.3 s OR 8.35 s HOWEVER, we kow that time cannot be negative so your correct answer is 8.35 seconds. Now, since we have the time for ty and tx=ty, we can solve for the distance that the motercycle makes: dx= (V1x)(t) + (0.5)(ax)(t)2 [remember, squared] dx= (24.75)(8.35) + (0.5)(0)(8.35)2 (now, since there is a zero, the second part cancels out and we get: dx= (24.75)(8.35) dx= 206.67m SO therefore, scott and jane WOULD make the jump on their harley. These projectile motion questions are a pain ;) Just practice lots and you'll get the hang of it. Good luck!

- Wood motorcycle ramp?
I am building a ramp for my harley. It is a touring bike and the dry weight is around 800 lbs. The wood I have cut is pressure treated pine. It is 4 feet wide and 3 feet long. The shed I am putting the bike in is about 10 inches high. Is this wood strong enough to hold that weight?

A Three feet to raise the bike ten inches is not much of a span. If you are just going to use it in the spring and fall to get your bike out and put it away again, you probably wouldn't have to brace a piece of 5/8" plywood in those dimensions. If you want it to hold up under repeated use, you should reinforce it with some 2x4 stringers. I built a 4x8 ramp to get up about 2 1/2' into my barn, reinforced it thus, and it is still in place 30 years later. Rolled all kinds of motorbikes and lawn tractors in and out. Gotta be careful, though; it's slippery when wet.