Kinematics question on a motorcycle

[imy786]

Homework Statement

A 100m long ramp is constructed perpendicular to a river’s edge with the
highest point 20 m directly above the edge of the bank
motorcycle stunt rider uses the ramp in an attempt to leap across a 30m wide
river. The motorcycle accelerates from rest at the bottom of the ramp and takes
off with a speed of 20 ms−1 moving parallel to the face of the ramp and
perpendicular to the river bank.

(a) What is the maximum height above the river banks reached by the
motorcycle?
(b) How long after take-off does the motorcycle land on the opposite bank?
(c) Calculate how far from the river the motorcycle lands.
(d) Calculate the velocity of the motorcycle at the moment of landing. Give your
answer both as the magnitude and direction of the velocity vector and in
component form.

Homework Equations

s= ut+1/2 at^2

v2= u2 + 2as

v= u+at

The Attempt at a Solution

(a)

(a)

arcsin0.2= 11.5369 degrees

u= 20 sin 11.5369...= 3.9458...

components in y diection-
s=?
v=0
u=3.9458...
a= -10

v2= u2+2as
o=15.569 + 2(-10) s
20s= 15.569
s= 0.77849 meteres...which doesn't seem correct...?

can anyone do this ??

 

[hage567]

Have you tried to calculate anything? You must show some work.
Hint for a): What can you say about the velocity of the rider when he reaches maximum height?

 

[imy786]

(a)
at max height, v=0, v2= u2 + 2as
0=400+ 2 (10) s
-400=20s
s=20 meters
??not sure

 

[hage567]

No, that's not quite right. Since the rider is leaving the ramp at an angle, you must break up the components of his initial velocity into the x and y directions. You've been given enough information in the question to find that angle.
Think about what is going on in each of these directions once he leaves the ramp. Write out the applicable kinematic equations for each direction so you can see how they fit together. v does equal zero at max height, but only for the y direction. Be careful with your signs, what is gravity doing to the rider as he reaches max height?

 

[imy786]

height=20
distance=30
30/20= 1.5
coz 1.5=?

components in y diection-
s=?
v=0
u=20
a=10

confused don't know what to do here..?

 

[hage567]

The 30 m is the width of the river though, right? What is the length of the ramp? Draw out the triangle so you can see how to solve for the angle. Once you have that, you can apply it to the velocity to get the components (hint: draw another triangle). Then put your new value for u (in the y direction) into your equation.

 

[imy786]

long ramp is constructed perpendicular to a river’s edge, this is kinda diagonal.


length of ramp=100m
inverse sin (20/100)=? give the angle= sin 0.2= ?

components in y diection-
s=?
v=0
u=20 inverse sin0.2
a=10

is this correct so far?

 

[hage567]

"inverse sin (20/100)=? give the angle= sin 0.2= ?"

This is backwards. If you take sin(theta)=20/100 from your triangle, then to find \theta you would apply the inverse function to get \theta = arcsin(20/100). Do you understand that part? So knowing that, how is it going to change

"u=20 inverse sin0.2"
?

 

[imy786]

yes by mistake i wrote sin 0.2
-----------------------------------
arc sin 0.2 = theta i understand.
this can be used for the velocity component.

u= 20 arcsin 0.2 this will give vertical inital velocty.

is that ok?

 

[imy786]

which part of the question have we done
and
are doing...?

 

[hage567]

You found that \theta=\arcsin0.2. What does this give you in degrees?

u= 20 arcsin 0.2 this will give vertical inital velocty.

OK, this is still not right, but try to work this out to get the actual magnitude for the velocity. Does the answer seem reasonable given what the total initial velocity is?
I'm not sure if you don't understand how to find the components, or are just getting confused with notation. I will show you what u should be:
By resolving the velocity vector you will find that for the y component, u=20\sin\theta. So you have two options here. One is to just put in the number in degrees in for \theta that I asked you to find above. The other is to put the expression for \theta into this, getting u=20\sin(\arcsin0.2)
If you cannot see why the sin should be there, I would suggest reviewing vectors/trig from your text or notes.
Work out the actual value for u. Does it seem reasonable?

which part of the question have we done
and
are doing...?

We haven't finished anything yet. Once you have this figured out, you can finish part (a) that you were working on before. This angle stuff is needed to do the question. This vector idea will also be needed again when you try to do part (d).

See what you can come up with for the rest of the problem.

 

[imy786]

(a)

arcsin0.2= 11.5369 degrees

u= 20 sin 11.5369...= 3.9458...

components in y diection-
s=?
v=0
u=3.9458...
a= -10

v2= u2+2as
o=15.569 + 2(-10) s
20s= 15.569
s= 0.77849 meteres...which doesn't seem correct...?

can anyone do this ??

 

[cristo]

(a)

arcsin0.2= 11.5369 degrees

u= 20 sin 11.5369...= 3.9458...

components in y diection-
s=?
v=0
u=3.9458...
a= -10

v2= u2+2as
o=15.569 + 2(-10) s
20s= 15.569
s= 0.77849 meteres...which doesn't seem correct...?

Why not? It seems correct to me. However, note that 20sin(theta) is exactly equal to 4 (since you worked out that sin(theta)=20/100)

 

[imy786]

But 0.7784 meteres doesn't it seem to less for a bike to go up after a traveling at a fast speed...i would assume it should go about 2-3 meters?

parts (b) (c) (d),,need help

 

[imy786]

if u= 4
then
components in y diection-
s=?
v=0
u=3.9458...
a= -10

v2= u2+2as
o= 16 + 2(-10) s
20s= 16
s= 0.8 meteres

(a) What is the maximum height above the river banks reached by the
motorcycle? 0.8 meteres.

 

[hage567]

(a) What is the maximum height above the river banks reached by the
motorcycle? 0.8 meteres.

The 0.8m is correct, but it is not the total height the motorcycle reached above the river. He jumped from a ramp that was above the ground, right?

 

[imy786]

b.

s= ut+1/2 at^2

if u= 4
then
components in y diection-
s=0.8
u=4
a= -10


s= ut+1/2 at^2
0.8 = 4t -5t^2

8= 40t-50t^2

4= 20t-25t^2

-25t^2 + 20t - 4 = 0

25t^2 - 20t + 4=0 solving quadratic equation

20 + 200 / 50 = 4.4sec or

20 - 200 / 50 = -3.6 secs

as time cannot be minus therefore t= 4.4 secs

 

[imy786]

total height = 20 + 0.8=
20.8 meters...

what about the rest of my answer part b.

 

[hage567]

b.

s= ut+1/2 at^2

if u= 4
then
components in y diection-
s=0.8
u=4
a= -10


s= ut+1/2 at^2
0.8 = 4t -5t^2

8= 40t-50t^2

4= 20t-25t^2

-25t^2 + 20t - 4 = 0

25t^2 - 20t + 4=0 solving quadratic equation

20 + 200 / 50 = 4.4sec or

20 - 200 / 50 = -3.6 secs

as time cannot be minus therefore t= 4.4 secs

Why did you pick 0.8 m?? Is he 0.8 m off of the ground when he leaves the ramp? Think about the equation and about parabolic motion. You will also have to be careful with your signs in this (depending on which way you select positive/negative displacement).

 

[imy786]

The 0.8m is correct, but it is not the total height the motorcycle reached above the river. He jumped from a ramp that was above the ground, right?

thats the reason i chose 0.8 meters.

im a bit confused now, can u help

 

[hage567]

The 0.8m is correct, but it is not the total height the motorcycle reached above the river. He jumped from a ramp that was above the ground, right?

thats the reason i chose 0.8 meters.

im a bit confused now, can u help

Think about what the question is asking. You need to find the total time from take-off to landing, right? Does the total time he is in the air equal the time it takes him to get 0.8m above the ramp? No, it does not.
So, what height above the ground is he when he starts the jump? What height at the end? You are looking for the change in vertical distance (Did you draw a picture and label everything? Because this distance is given in the problem you don't need to calculate it). Put that into your equation and solve for t. Like I said before, you will need to be careful with the sign you give to this change in distance.

Note: In post #17 when you first tried to solve this with the 0.8m, you did not solve the quadratic equation properly. Also, you don't need to multiply through to get whole numbers (unless you just like it that way).

 

[imy786]

(a) What is the maximum height above the river banks reached by the
motorcycle?

total height = 20 + 0.8=
20.8 meters...

is this corect?

 

[cristo]

(a) What is the maximum height above the river banks reached by the
motorcycle?

total height = 20 + 0.8=
20.8 meters...

is this corect?

Yup, looks fine.

 

[imy786]

cristo, why do you think hage567 ,

Why did you pick 0.8 m?? Is he 0.8 m off of the ground when he leaves the ramp? Think about the equation and about parabolic motion. You will also have to be careful with your signs in this (depending on which way you select positive/negative displacement).

-------------------------------------------------------

 

[imy786]

cristo...i used v=0 to work out height above the river as 0.8 meters.
But at max height v isn't 0 is it?

coz the motorcyle is still traveling in the air at max hight.

 

[cristo]

cristo, why do you think hage567 ,

Why did you pick 0.8 m?? Is he 0.8 m off of the ground when he leaves the ramp? Think about the equation and about parabolic motion. You will also have to be careful with your signs in this (depending on which way you select positive/negative displacement).

-------------------------------------------------------

I'm not sure; you'll have to ask him! I remember checking this question last week, and came up with 0.8m

cristo...i used v=0 to work out height above the river as 0.8 meters.
But at max height v isn't 0 is it?

coz the motorcyle is still traveling in the air at max hight.

You set the vertical velocity equal to zero-- this is correct. Whilst the bike is still moving at the maximum height, its velocity is soley in the horizontal direction.

 

[hage567]

cristo, why do you think hage567 ,

Why did you pick 0.8 m?? Is he 0.8 m off of the ground when he leaves the ramp? Think about the equation and about parabolic motion. You will also have to be careful with your signs in this (depending on which way you select positive/negative displacement).

-------------------------------------------------------

Why do you think hage567 what?? I don't understand what you are trying to say.

 

[imy786]

hage don't worry...leave it...probly my misunderstanding..

lets carry on with the quesition.

(b) How long after take-off does the motorcycle land on the opposite bank?

to work out the time it takes for the cycle to reach the bank. It would been
-20metres reletive to the start off the ramp.

so is it ok to assume
s= -9.8
a= -10
u= 4
t=?
s= ut+1/2 at^2
to use this method to work out t is that corect

 

[harmeet_angel]

A 100m long ramp is constructed perpendicular to a river’s edge with the
highest point 20 m directly above the edge of the bank
motorcycle stunt rider uses the ramp in an attempt to leap across a 30m wide river

I haven't gone through every single reply... but wondering if that's really possible that 100m ramp constructed over a river with a width of 30m ??
I am confused maybe wrong

 

[cristo]

hage don't worry...leave it...probly my misunderstanding..

lets carry on with the quesition.

(b) How long after take-off does the motorcycle land on the opposite bank?

to work out the time it takes for the cycle to reach the bank. It would been
-20metres reletive to the start off the ramp.

so is it ok to assume
s= -9.8
a= -10
u= 4
t=?
s= ut+1/2 at^2
to use this method to work out t is that corect

No. Firstly, you need to split the motion into horizontal and vertical components. Try again. One hint though; the total displacement in the y direction is zero (since we presume that the banks are equal heights on either side of the river.)

 

[hage567]

I haven't gone through every single reply... but wondering if that's really possible that 100m ramp constructed over a river with a width of 30m ??
I am confused maybe wrong

It's not over the river, it's leading up to the edge of the river.

 

[harmeet_angel]

oh yea, thanks.

 

[imy786]

s= 0
a= -10
u= 4
t=?
s= ut+1/2 at^2
0= 4t -5t^2
4t=5t^2
4/5= t= 0.8 seconds

is this correct

 

[hage567]

If this is to find the time from when he leaves the ramp to when he lands on the other side of the river, then NO, it isn't correct. s should not be zero, since the top of the ramp is above the ground. What is his vertical displacement from the ramp to the ground?

 

[hage567]

No. Firstly, you need to split the motion into horizontal and vertical components. Try again. One hint though; the total displacement in the y direction is zero (since we presume that the banks are equal heights on either side of the river.)

Ah, I see why you may have thought that, imy786. In this post, cristo says the displacement is zero since the banks are equal height. But this is not the case, since he is jumping from the top of the ramp. So put the correct number in for s, and you should get the correct answer.

 

[imy786]

well if S is not 0, what is it then?
iv put s as -9.8 before that is incroct as well..

need help??

 

[hage567]

s represents the vertical displacement. 9.8 is the acceleration due to gravity. They are not the same thing.
Here's some advice: keep track of the units of all your numbers. You must keep your variables straight. Try to think about the equations you are using and what they mean. Randomly guessing what numbers to put in is only going to cause more confusion.

So, if the height of the ramp above the ground is 20 meters, what is his vertical displacement when he lands on the ground on the other side of the river? That is what you need to put into your equation. See what you get now.

 

[imy786]

-20 is displacemet

is that right

 

[hage567]

Yes! That is right.

 

[imy786]

s= - 20
a= -10
u= 4
t=?
s= ut+1/2 at^2
-20= 4t-5t^2
4t-5t^2+ 20= 0
solving this quadrtic will give value for two time.

 

[hage567]

Yes, but you should get one positive root and one negative root. Which one do you think you should use?

 

[imy786]

(c) Calculate how far from the river the motorcycle lands.
[need help on this part]

i think we have to take horizontal motion into consideration-
s= ?
u= 20 cos 11.53...= 19.6
a= 0 (horizontal no accelration i think)
t= (do i use the t from previos answer (b) ]

then using this equation of motion

s= ut+1/2 at^2

is this correct?

 

[imy786]

urgent help needed...please help me do the quesion

 

[cristo]

(c) Calculate how far from the river the motorcycle lands.
[need help on this part]

i think we have to take horizontal motion into consideration-
s= ?
u= 20 cos 11.53...= 19.6
a= 0 (horizontal no accelration i think)
t= (do i use the t from previos answer (b) ]

then using this equation of motion

s= ut+1/2 at^2

is this correct?

This is the correct method, presuming that the time you calculated in part b is correct.

 

[imy786]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.

taking horizontal motion-

v=?
u=
a=o
s=

but if we use v2= u2+ 2as
final velocity would equail intital velocity by calculation (if a=0)

need help?

 

[cristo]

No, the angle will be different. You need to set up two equations for this; one in the horizontal, one in the vertical.

 

[imy786]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------------------------------
2 equations- horizontal and verical motion-

horizontal motion -

v=
u=?
a=0
s=

what would s and v be for horiznotal motion...

 

[hage567]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------------------------------
2 equations- horizontal and verical motion-

horizontal motion -

v=
u=?
a=0
s=

what would s and v be for horiznotal motion...

You tell me what you think they should be.

You know the initial horizontal velocity. If there is no acceleration in the horizontal direction, what would it be at the end of the jump?

You can easily find s, because you know the total time of flight (you found that already). So how do you find distance if you know the time and the velocity?

 

[imy786]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------------------------------
2 equations- horizontal and verical motion-

horizontal motion -

v=
u=?
a=0
s= 10 m (eg assuming this is the value from b)
t= 2 (leg ets assume we calculated t as 2 secs)
if there is no acceratlion then u=V

 

[hage567]

(d) Calculate the velocity of the motorcycle at the moment of landing. Give your answer both as the magnitude and direction of the velocity vector and in
component form.
--------------------------------------------------------------------------
2 equations- horizontal and verical motion-

horizontal motion -

v=
u=?
a=0
s= 10 m (eg assuming this is the value from b)
t= 2 (leg ets assume we calculated t as 2 secs)
if there is no acceratlion then u=V

Why are we assuming anything? I'm confused as to what you really trying to find. You know what t is from part b. I don't know where the 10 m comes from, I don't see it in any previous post. Isn't s what you were supposed to find for part c? Did you do that? That's the horizontal distance, right?

You don't need s it to do part d anyway. What is the vertical component of velocity at landing? That's what you should be working on, since you already know the horizontal component.