How can you accurately hit a trash bin while biking with an apple core?

[Keeptrying83]

Homework Statement

You are traveling on a bike, moving at a constant speed of 20,0 km/h. You just finished eating an apple, and you wish to pop the core into a trash bin located by a busstop.The bin is 3,00 m from the edge of the road (where you are biking). You throw it at a point 4,00 m before you pass the trash bin. The apple core behaves as a projectile under the force of gravity g = 9,80 m/s^2, and you can ignore friction and wind. The height of the (top of the) bin is the same as the height of your hand as you throw, 1,00
m above the ground.

a) You throw the apple upwards at an angle of = 20 degrees with horizontal. With what initial velocity vector relative to the Earth should you throw the apple to hit the bin?

b) With what initial velocity vector relative to you and the bike should you throw the apple to hit the bin? What speed does that correspond to?

c) As it happens, a little old lady is standing exactly halfway between you and the bin. She is 160 cm tall. Do you hit her? Remember to draw a sketch of the situation.

Homework Equations

For (a) I used the range equation, R=(v^2 * sin(thetha))/g

The Attempt at a Solution

For (a) I used the range equation R=(v^2 * sin(thetha))/g and got 8,73 m/s.

In (b) I struggle with getting started. If the bike relative to the Earth is the velocity vector in y-direction (20 km/h), and the the apple relative to Earth is the second vector. Then the velocity vector for me relative to the bike is the opposite vector from the angle? I am really stuck on this one, hope someone can help me here:)

 

[RUber]

Could you solve the problem if the bike was not moving? If you can, just subtract the velocity of the bike from your stationary solution.

 

[RUber]

I concur that the total velocity in the direction of travel is what you found for part (a), |V|=8.73m/sec.
What portion of that is in the +z direction? What portion is in the xy plane? Of the portion in the xy plane, what are the x and y components?
Subtract off the velocity of the bike from the y component and you should have a solution.

The question about hitting the lady can be reduced to the 1-dimensional problem based on the initial upward velocity and time to the middle of the path.

 

[Sabalaba]

how did you get 8,73 m/s on a)?

 

[hefalomp]

Have you managed to do b or c?

 

[RUber]

how did you get 8,73 m/s on a)?

I agree that the range equation as listed :

The Attempt at a Solution

For (a) I used the range equation R=(v^2 * sin(thetha))/g and got 8,73 m/s.

Does not lend itself right away to finding 8.73 m/sec.

However, the range equation can be derived by using two pieces of information:
1) The total change in z is zero. $R_z = vt_f\sin\theta - \frac{g}{2}t_f^2 = 0$.
Solving for t, you get the trivial solution of t=0, or $t_f = \frac{2v}{g}\sin \theta$
2) The total distance in the x-y plane $5m = R_{xy}= v t_f \cos\theta$. Replacing t_f with what you found above gives:
$5m =v\frac{2v}{g}\sin \theta \cos\theta =\frac{2v^2}{g}\sin \theta \cos\theta$
Rearranging to solve for v, you get:
$v = \sqrt{ \frac{5g}{2 \sin \theta \cos \theta } }$
Which gives v = 8.731m/sec.

 

[hefalomp]

I agree that the range equation as listed :

Does not lend itself right away to finding 8.73 m/sec.

However, the range equation can be derived by using two pieces of information:
1) The total change in z is zero. $R_z = vt_f\sin\theta - \frac{g}{2}t_f^2 = 0$.
Solving for t, you get the trivial solution of t=0, or $t_f = \frac{2v}{g}\sin \theta$
2) The total distance in the x-y plane $5m = R_{xy}= v t_f \cos\theta$. Replacing t_f with what you found above gives:
$5m =v\frac{2v}{g}\sin \theta \cos\theta =\frac{2v^2}{g}\sin \theta \cos\theta$
Rearranging to solve for v, you get:
$v = \sqrt{ \frac{5g}{2 \sin \theta \cos \theta } }$
Which gives v = 8.731m/sec.

I got that v = sqrt((g*5)/(2(tan(theta)*cos^2(theta)))
which also gives 8.73m/s, do you think this is the correct answear then?

 

[RUber]

tan x cos^2 = sin x cos.

 

[coffeemanja]

May in b it is just that simple as to covert 20km/h to m/s, then subtract the result from the answer in a?

 

[hefalomp]

May in b it is just that simple as to covert 20km/h to m/s, then subtract the result from the answer in a?

So you think if the velocity should add up to be 8,73m/s as if the bike was stationary relative to the earth, the velocity relative to you and the bike should be 8,73=(20km/h)/3.6 * v(youandbike), and solve for v(youandbike) = 3.17m/s ?

 

[hooolahooop]

In a), don't you need an angle in your answer because they are asking for a vector? know its relative to the earth, therefore i am not exactly sure if you need an angle in your answer

 

[RUber]

I assume that the problem is asking for the vector in component form.
$\vec V = V_x \hat x + V_y \hat y+V_z \hat z$.
The velocity of the bike is pointed in my +y direction, so in vector form,
$\vec V_{bike} = 0 \hat x + V_{bike} \hat y+0 \hat z$, where
$V_{bike}= | \vec V_{bike} |$
For part b, the vector describing how you throw the apple is exactly $\vec V - \vec V_{bike}$
But the magnitude of that vector is not $|\vec V| - |\vec V_{bike}|$, since you can only directly subtract magnitudes if the vectors are pointing in the same direction.

 

[jensjensen]

Is this correct method on b?
From bicycle:
V_xi=5.56 m/s

From throw
V_xi=V_i*cos(36.7)
V_yi=V_i*sin(20)
V_zi=V_i*cos(53.3)

Then i find the velocity in yz-direction using V_yi=1/2*g*t, where t=5/(V_zi*cos(53.3)), and velocity in xy-direction using V_yi=1/2*g*t, where t=5/(V_xi*sin(36.7)). Add the V_yi-velocities and subracts the velocity from bicycle in V_xi-direction.
And then i take sqrt(V_xi²+V_yi²+V_zi²). That makes me end up with 9.77 m/s.

 

[RUber]

Is this correct method on b?
From bicycle:
V_xi=5.56 m/s

From throw
V_xi=V_i*cos(36.7)
V_yi=V_i*sin(20)
V_zi=V_i*cos(53.3)

Then i find the velocity in yz-direction using V_yi=1/2*g*t, where t=5/(V_zi*cos(53.3)), and velocity in xy-direction using V_yi=1/2*g*t, where t=5/(V_xi*sin(36.7)). Add the V_yi-velocities and subracts the velocity from bicycle in V_xi-direction.
And then i take sqrt(V_xi²+V_yi²+V_zi²). That makes me end up with 9.77 m/s.

Nope.
You need to keep a cos(20) term in your Vx and Vz numbers. Otherwise your sum of squared components will not add up to Vi^2.

 

[hooolahooop]

Is this correct method on b?
From bicycle:
V_xi=5.56 m/s

From throw
V_xi=V_i*cos(36.7)
V_yi=V_i*sin(20)
V_zi=V_i*cos(53.3)

Then i find the velocity in yz-direction using V_yi=1/2*g*t, where t=5/(V_zi*cos(53.3)), and velocity in xy-direction using V_yi=1/2*g*t, where t=5/(V_xi*sin(36.7)). Add the V_yi-velocities and subracts the velocity from bicycle in V_xi-direction.
And then i take sqrt(V_xi²+V_yi²+V_zi²). That makes me end up with 9.77 m/s.

How did you get the angles 53.5 and 36.7?

 

[jensjensen]

36.7 is the angle between where you start throwing and the red arrow. 53.3 is 90-36.7. But should probably not use them.

 

[RUber]

The angles are fine. Notice that you can keep a more precise answer by saying that the right/left component (x in my diagram) is 3/5 the total 2d velocity, and the front/back component (y in my diagram) is 4/5 the total 2d velocity.
Otherwise, cos(53.3) = .5976, cos(36.7) = .8018. These are probably close enough to not affect your rounding...but why go through the effort of finding the angles when the side lengths are given to you and the hypotenuse is simple enough to figure out?

 

[tom ryen]

Vxi=Vi*cos(36.7)*cos(20)=6.563 m/s
Vyi=Vi*sin(20) = 2.986 m/s
Vzi=Vi*cos(53.3)*cos(20)= 4.922 m/s

Vxrel=Vxi - Vbike= 6.563 - 50/9 = 1.007 m/s

Virel = Vxrel / cos(36.7)*cos(20)= 1.007/ cos(36.7)*cos(20)= 1.18 m/s

is 1.18 m/s the initial velocity vector relative to you and the bike should you throw the apple to hit the bin?

 

[RUber]

1.18 m/sec is a speed. And no.
I agree with what you did up to ...

Vxi=Vi*cos(36.7)*cos(20)=6.563 m/s
Vyi=Vi*sin(20) = 2.986 m/s
Vzi=Vi*cos(53.3)*cos(20)= 4.922 m/s

Vxrel=Vxi - Vbike= 6.563 - 50/9 = 1.007 m/s

there.
But remember, your throw is responsible for all of the other components.

 

[insightful]

Vxi=Vi*cos(36.7)*cos(20)=6.563 m/s
Vyi=Vi*sin(20) = 2.986 m/s
Vzi=Vi*cos(53.3)*cos(20)= 4.922 m/s

Vxrel=Vxi - Vbike= 6.563 - 50/9 = 1.007 m/s

Looks good. For Virel, check your answer by first calculating time of flight, t, and see if 1.18m/s "gets you there."

 

[tom ryen]

hmm..
ifI understand you right RUber I should get it by this formula : sqrt(Vxrel^2+Vyi^2+Vzi^2) ?

 

[RUber]

hmm..
ifI understand you right RUber I should get it by this formula : sqrt(Vxrel^2+Vyi^2+Vzi^2) ?

That will give you the magnitude of your velocity (i.e. speed).
Most vectors are expressed in their component form, which you already have.

 

[jensjensen]

So the thing i did now was that is that i changed z=3 when i looked at yz-vectors and x=4 when i looked at xy-vectors

From bicycle:
Vxi=5.56 m/s

From throw
Vxi=Vi*cos(20)
Vyi=Vi*sin(20)
Vzi=Vi*cos(20)

Then i find the velocity in yz-direction using Vyi=1/2*g*t, where t=(3/(Vzi*cos(20)), and velocity in xy-direction using Vyi=1/2*g*t, where t=4/(Vxi*cos(20)). Add the Vyi-velocities and subracts the velocity from bicycle in Vxi-direction aka V_relxi=7.81*cos20-5.56=1.78, V_yi=7.81*sin(20)+sin(6.76)=4.98 and V_zi=6.76*cos(20)=6.35
And then i take sqrt(Vrelx2+Vyi2+Vzi2). That makes me end up with 8.27 m/s.

 

[kongen]

so is that the right answer?

 

[coffeemanja]

So you think if the velocity should add up to be 8,73m/s as if the bike was stationary relative to the earth, the velocity relative to you and the bike should be 8,73=(20km/h)/3.6 * v(youandbike), and solve for v(youandbike) = 3.17m/s ?

Yes

 

[RUber]

Yes

That's not right.
You cannot directly subtract the magnitudes of the velocity from each other. This only works when the velocities are pointing in the same direction.

The total velocity of the apple when it starts its journey is 8.731m/sec. And it is pointed at a 20 degree elevation, right toward the trash can. In the x-y plane, it will have a straight line course, in the plane drawn between z and the direction of travel, it will have a standard parabolic ballistic trajectory.

You can answer part C based on the initial angle and initial velocity.
For part B, you need to break the velocity into components, x- component, y-component, z-component. Double check your components with
$v_x ^2 + v_y^2 + v_z^2 = V^2$.
Once you have components, you can directly subtract the velocity from the bike -- component-wise. That way you are only dealing with velocities in the same directions.

Review the posts in this thread. There have been at least a couple people who have posted the correct answers...and quite a few who have posted incorrect methods.

 

[coffeemanja]

That will give you the magnitude of your velocity (i.e. speed).
Most vectors are expressed in their component form, which you already have.

So, in b) if I have vectors Vi=8,73, Vb(bike)=5,6 (that is m/s from 20 km/h) and Va...Va=3,13.
To find initial velocityof the throw relative to me on the bike sqrVb+sqrVa=sqrx, X=6,4m/s.
Relative to me on the bike, displacement is 3 meters...then I can find time t=3/6,4=0.5 sec
The speed will be: dist. traveled by the apple (5m)/time (0,5s)=10 m/s

 

[RUber]

So, in b) if I have vectors Vi=8,73, Vb(bike)=5,6 (that is m/s from 20 km/h) and Va...Va=3,13.
To find initial velocityof the throw relative to me on the bike sqrVb+sqrVa=sqrx, X=6,4m/s.
Relative to me on the bike, displacement is 3 meters...then I can find time t=3/6,4=0.5 sec
The speed will be: dist. traveled by the apple (5m)/time (0,5s)=10 m/s

What is Va? Are you still subtracting the velocity of the bike directly from the velocity of the apple? How can the speed you calculated at the end be more than the speed you started with of 8.73? That is not right.

First, break the initial velocity into its components.

8.731 * sin(20) is your vertical component
8.731 * cos(20) is your flat component.
3/5 ( 8.731 * cos(20) ) is the portion of the flat component moving to the *edit* right.
4/5 ( 8.731 * cos(20) ) is the portion of the flat component moving forward.

Then, and only then, can you subtract the velocity of the bike. And, you only subtract the velocity of the bike from the component pointing in the same direction as the bike.
Once you have done that, you have your answer...in vector component form.

If you want to see the speed, then take sqrt ( [8.731 * sin(20)]^2 + [3/5 ( 8.731 * cos(20) ) ]^2 + [ 4/5 ( 8.731 * cos(20) ) - Vbike ] ^2 ).
That is the speed in the direction of the throw.

 

[coffeemanja]

What is Va? Are you still subtracting the velocity of the bike directly from the velocity of the apple? How can the speed you calculated at the end be more than the speed you started with of 8.73? That is not right.

First, break the initial velocity into its components.

8.731 * sin(20) is your vertical component
8.731 * cos(20) is your flat component.
3/5 ( 8.731 * cos(20) ) is the portion of the flat component moving to the left.
4/5 ( 8.731 * cos(20) ) is the portion of the flat component moving forward.

Then, and only then, can you subtract the velocity of the bike. And, you only subtract the velocity of the bike from the component pointing in the same direction as the bike.
Once you have done that, you have your answer...in vector component form.

If you want to see the speed, then take sqrt ( [8.731 * sin(20)]^2 + [3/5 ( 8.731 * cos(20) ) ]^2 + [ 4/5 ( 8.731 * cos(20) ) - Vbike ] ^2 ).
That is the speed in the direction of the throw.

Va is a relative velocity. http://www.physicstutorials.org/home/mechanics/1d-kinematics/relative-motion

 

[coffeemanja]

I just do not get why is there portions moving to the left, as you say...

 

[RUber]

Va is a relative velocity. http://www.physicstutorials.org/home/mechanics/1d-kinematics/relative-motion

Thank you for clarifying.
In any case, you did not add your vectors properly.
Va + Vb = Vi, right? And Vi has vertical, right, and forward components. Vb only has a forward component.
So Va will have the same vertical and right component as Vi. Va's forward component will be Vi(forward) - Vb.

 

[RUber]

I just do not get why is there portions moving to the left, as you say...

Edited, that was moving to the right in the diagram.

 

[kongen]

What is Va? Are you still subtracting the velocity of the bike directly from the velocity of the apple? How can the speed you calculated at the end be more than the speed you started with of 8.73? That is not right.

First, break the initial velocity into its components.

8.731 * sin(20) is your vertical component
8.731 * cos(20) is your flat component.
3/5 ( 8.731 * cos(20) ) is the portion of the flat component moving to the *edit* right.
4/5 ( 8.731 * cos(20) ) is the portion of the flat component moving forward.

Then, and only then, can you subtract the velocity of the bike. And, you only subtract the velocity of the bike from the component pointing in the same direction as the bike.
Once you have done that, you have your answer...in vector component form.

If you want to see the speed, then take sqrt ( [8.731 * sin(20)]^2 + [3/5 ( 8.731 * cos(20) ) ]^2 + [ 4/5 ( 8.731 * cos(20) ) - Vbike ] ^2 ).
That is the speed in the direction of the throw.

where do you get the angle 20 from?

 

[coffeemanja]

Thank you for clarifying.
In any case, you did not add your vectors properly.
Va + Vb = Vi, right? And Vi has vertical, right, and forward components. Vb only has a forward component.
So Va will have the same vertical and right component as Vi. Va's forward component will be Vi(forward) - Vb.

Yes, if you look at the diagram as at is in the problem...origin is the bike, x to the right, y -up. The Vi forward component is equal 8,73...no?

 

[coffeemanja]

If we look at the situation as flat diagram, vi(flying apple) is 8,73. Vb is 5,6. Given that relative velocity = v object (Apple)-v observer(me on the bike). I find angles irrelative.