physics help on initial velocity, have an exam tomorrow?

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).

the ramp is 12m long and the angle is 30 degrees but fill in any figures its more the method im looking for

Answer 1

I just saw this problem. Hope it's not too late for your test. Rico's answer is wrong; amandeep's is right, but he made a bad assumption (45 degrees) and didn't show his method.

The initial velocity is v. It has two components -- vh (horizontal) and vv (vertical) -- where

vh = v cos 30 = v sqrt(3) / 2
vv = v sin 30 = v/2

Once the bike is airborne, it will stay airborne until it lands. Since the horizontal velocity is constant,

sh = vh t where sh (horizontal distance) is 100 m.
100 = [v sqrt(3) / 2] t
t = [200 sqrt(3)] / (3v)

That's how long it must remain in the air, in seconds.

Notice that t^2 = 40000 / (3v^2)

Now do the vertical part, using the formula

sv = (1/2) a t^2 + vv t + so

where sv is vertical distance; a = g = -9/8 m/s/s (gravity); and so is initial vertical displacement.

In this problem, so = 12 sin 30 = +6 meters (top of the ramp); and sv = 0. (The bike ends up at ground level.)

Plug in some values to get

0 = (1/2) (-9.8) t^2 + (v/2) t + 6

Now plug in the t & t^2 values to get

0 = (-4.9)(40000) / (3v^2) + (v/2) [200 sqrt(3)] / (3v) + 6

(4.9)(40000) / (3v^2) = [100 sqrt(3) + 18] / 3

v^2 = (4.9)(40000) / [100 sqrt(3) + 18]

v^2 = 1025.077363

v = 32.0168 m/s

which rounds off to v = 32 m/s

which happens to be the same answer amandeep got.

Answer 2

I think I can help Or hope that I can
For simplicity sake I m taking angle of the ramp as 45 degrees
Now we know that the horizontal distance travelled by a projectile is given by u^2sin2A/g
where A is angle of projection
From this eqn. we calculate the value of u reqired to cross the Canyon It comes out to be around 31.305 m/s Therefore the stuntman must leave the ramp at the velocity of at least 32m/s
I think this must be the solution or at least I hope it is

Answer 3

I think you have to give us the angle of the ramp in relation to the ground before you can calculate.

Answer 4

The initial speed has to be enough to obtain a horizontal range of 112 meters.

There is velocity in two directions

vx(0)
vy(0)

Where the overall velocity is defined as:

v(0) = vx(0) / cos 30
or
v(0) = vy(0) / sin 30

For the horizontal range:

y = vy(0)*t - 4.9t^2
At the end of the jump, y= 0 and

0 = vy(0)*t - 4.9t^2
vy(0)*t = 4.9t^2
t = vy(0) / 4.9

Also,
x = vx(0)*t
So, x = [vx(0) * vy(0)] / 4.9
But x is just horizontal range:

112 = [vx(0) * vy(0)] / 4.9
[vx(0) * vy(0)] = 548.8

and from above
v(0) = vx(0) / cos 30
v(0) = vy(0) / sin 30

Multiply the two equations:

[v(0)]^2 = vx(0)*vy(0) / [cos 30 * sin 30]
[v(0)]^2 = 548.8 / [cos 30 * sin 30]
[v(0)]^2 = 1267.4
v(0) = 35.6 m/sec * 3600 sec/hour * .000621 mile/m
=79.6 mph

Answer 5

Geeeee...... If we can't see the drawing, it's gonna play merryhell with our calculations.

'Course, we won't be able to see the drawings on your exam tomorrow, either.............................


Doug