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A physics professor did daredevil stunts in his spare time. His last stunt was an attepmt to jump accross a river on a motorcycle. The takeoff ramp was inclined at 53 degrees, the river was 4 meters wide, and the far bank was 15 meters lower than the top of the ramp. The river itself was 100 meters below the ramp. Ignore air resistance.
a.) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?
b.) If his speed was only half the value found in (a), where did he land?

2006-08-07 01:04:19 · 4 answers · asked by just_askin' 1 in Science & Mathematics Physics

tnx for the answer, but i can'y understand what you wrote.. you stopped at:

a)vcos53*t=4 , t=vsin53/g,v^2sin106/2g=4,v^2=...

is there a next to this? hope you can make me understand... i'm really having a hard time.. sigh...

2006-08-07 05:00:11 · update #1

4 answers

good question.i am just going to explain you the procedure you can work it out.
Diagram will be somthing like this:
... e
.../|.....f
../ |.....|\
/----|-----|--\
a...b...c...d


dots means spaces.The diagrams are not comming as i wanted them to.i suggest u drawn them in your note book


angle bae is 53
eb=100m
fc=100-15=85m
bc=4m

now heres the trick . u gotta use concept of similar triangles also. the diagram becomes like this
e
..../|......f
h /--|g----|\
/----|------|--\
a....b.....c..d
i know it aint accurate but its just to give u an idea.
now eg is 15m
and angle ghe = 53
so your diagram becomes
e
.../| ......f
h/--|g---- |\(actually f would be in the same level as g)

Now that the question is simplified throught diagram its become very simple.
use trignometry (tan53=eg/hg) to find hg
we have to find min vel so that the *** err proff reaches f.
for this we can use range concept: R=(u^2)sin2x/g
R is hg+gp=hg+4
find u using the above formula
now use vertical component eqn v^2 = u^2 -2gY (where Y is 15m)
find the vertical component of u (usinx) and put in above eqn.
get the value of v.This is your vertical comp velocity.
Find the horizontal component of velocity (which will always be ucosx at any point of the projectile).
Add the horizontal and vertical comp vectorially (just take the root of the sum of squares ie root([hor vel]^2 + [ver vel]^2))
The answer you get must be the vel he should have at the top of the ramp.

(B PART)Half this velocity and find the inetial velocity corresponding to this velocity (just go back/reverse the above procedure) put this inetial velocity in the range equation and find the range.subract hg from this range and that will be the answer.

2006-08-07 06:34:16 · answer #1 · answered by Anonymous · 0 1

This is not hard. Trust me. Just requires a bit of algebra.

What you will end up with is an equation looking something like this:

[((v cos (53))/4)(v sin (53))] -9.8[(v cos (53))/4]^2 = 15

Solve for v for the first question, half it and use your kinimatics equations to find out where he smacks into the cliff walls in the second one. Fun question, but took me longer to type out than it should have taken you to solve.

2006-08-14 22:19:37 · answer #2 · answered by Roger N 2 · 1 0

The problem is a simple kinematic exercise. I won't solve the problem for you, but I will tell you that if you google "Kinematic" you will find all the formulas you need.

2006-08-14 10:11:27 · answer #3 · answered by sparc77 7 · 0 2

a)vcos53*t=4 , t=vsin53/g,v^2sin106/2g=4,v^2=80/sin106 if you give me your email id then i will mail you here it is difficult to explain my id is"disco5z@yahoo.com"

2006-08-07 08:20:26 · answer #4 · answered by disco5z 1 · 0 1

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