physic of mechanics i need major help i just cant grasp the method here?

Question

A film director is organising a stunt for his latest film. A stuntman will jump over part of the Grand Canyon on a motorbike. The canyon is only 100 m wide at the part where the ramp has been built and the length of the ramp is 12 m. As a safety inspector you have been asked to assess if the jump is possible. Using the diagram of the ramp below you must calculate the minimum speed the motorbike must by doing at the top of the ramp if it is to make it across. (You can ignore air resistance or wind).

ramp is 30 degrees so opp is 6m

Answer 1

I answered this in the other thread, so I'll explain it here. (The algebra is in the other thread.)

First, DG's answer is wrong because he ignored the 6-meter height (initial vertical displacement) of the ramp. He assumed you begin and end at the same height. Bad assumption.

Now I'll just talk about method. From the ramp geometry (12 meters, 30 degrees), you know he starts with an initial vertical displacement of 6 meters (12 sin 30). You also know his initial velocity has a 30 degree angle. So if v (or vo if you prefer) is initial velocity, then the horizontal and vertical components are vh = v cos 30 and vv = v sin 30.

Do the horizontal part next. You know it has to go 100 meters, and, since horizontal velocity is constant, 100 = vh t, where t is time in the air. Solve this for t = f(v) for use in the vertical portion of the problem.

Now do the vertical part. All you have to think about is going straight up and coming straight down; the horizontal part is already done.

The vertical displacement formula is s = (1/2)at^2 + vv t + so
where a = g = -9.8 (gravity); s is the ending displacement (zero meters of height when it lands across the canyon); so is the +6 meters of initial height.

Now you have everything you need to do this problem. Just plug in the values you have, including t = f(v), the time in the air, and you can solve it for v.

The answer, which you can see in the other thread, is 32 m/s/s (actually something like 32.01, but I thought 32 was close enough).

Hope this explanatory answer helps you understand what's going on. Splitting the problem into vertical and horizontal components is the key. And hope you do/did well on your test.

Answer 2

The length of the ramp is irrelevant as long as the ramp ends at the edge of the canyon.

Now, use the fundas of projectile motion and the distance covered by projectile
= 2 v^2 sinx cosx/ g

x = 30 deg, take g = 10 or 9.8 to be precise

so, 100 = 2 v^2 sin30 cos 30 / 9.8

v^2 = 1131.6

so, v = 33.6 m/sec

= 33.6 x18/5 Km/h

= 121 Km/h

Answer 3

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Answer 4

This is simple kinematics. Use your kinematic equations, you have all of your variables, start by labeling which variable is which. Because the ramp is there you will have an initial x and y velocity in which the magnitude will be the resulting velocity of the bike. Work backwards in a sense.

Answer 5

If we ignore air resistance, the stuntman will not make it to the other side of the canyon.