Motorcycles and physics 2

how fast is he going?

 

45 degrees. Wanted to be first in with the answer, so explanation follows on the edit.

The parameter we need to maximise is called range. It is the horiz distance travelled by a projectile describing a parabolic arc free falling under gravity between the point of launch and where it hits the ground.

Easiest way is to just let the particle have a launch velocity with a magnitude V and angled theta (t) to the horizontal. Resolve that into a vertical component Vsin t and a horiz component Vcos t. These two motions are independent, the vertical being affected by gravity (-g) towards the ground and the horiz unaffected.

When the object hits the ground again, it's vertical velocity (by conservation of energy) will have the same magnitude V only pointing downwards.

Using the elementary eqn v = u + at

-Vsin t = +Vsin t + (-g)T

T = (2Vsin t)/g

T being the time it takes for the object to hit the ground again.

The horiz displacement at time T is given by (distance = speed X time)

s_x = (2Vsin t/g)*(Vcos t) = (V^2sin 2t)/g using a simple trig double angle identity.

Differentiate s_x partially wrt theta (t). It's just the same as complete differentiation, except it's more general because it becomes independent of variable V. Since I can't denote "curly d" here, I'll just use 'd'

d/dt(s_x) = (2V^2/g)cos 2t

Set that to zero to find maxima/minima,

cos 2t = 0

t = 45 degrees.

Differentiate again to prove that this is a maximum.

d^2/dt^2(s_x) = - (4V^2/g)sin 2t

which is negative at t = 45 degrees. We're done now.

 

0.00000000000000000001 degrees will give you max distance (Not hight), 0 degrees is even longer but then it's not a "Jump".



(If you consider suspension travel and weight it might be a little more tricky but you didn't mention it).

 

Same as the angle you fire a bullet at to get max distance, 45 degrees.
(Ignoring wind resistance, etc.)

 

Hey, YOU said "using physics/math/trig/logic". I just tried to cram as much of all 4 as I could into there.

I'm a hopeless teacher, I'll freely admit it.

Thanks.

 

i think your explination is sufficient.

to calculate x and y values of a velocity vector, we have a couple of equations.
x = v*cos(angle btwn ground and velocity vector)
y = v*sin(angle)

to find the max x AND y values, you can logically think about it (using yellow360's method), you can go all out on the math like turboflat4, or you can whip out a graphing calculator and use a little math knowledge.

the cos funct. starts at (0,1) and decreases on the interval [0,pi]
the sin funct starts at (0,0) and increases on the interval [0,pi/2]
we'll cheat and assume that they will intersect somewhere. that intersection is the max of the sin AND cos functions at any given time.

therefore the intersection of sin(x) and cos(x) on [0,pi] should be the angle such that max distance is achieved. [we get .707. arcsin or arccos(.707) is 45*]

there i tried to explain it. but logically looking at it is the fastest and therefore IMO the best way to look at it