# Projectile Motion of motorcyclist

• gjh
In summary, the problem statement involves a motorcyclist driving in a 60-km/h zone who hits a stopped car and is thrown from his bike, landing 39 m down the road. The question is whether or not the motorcyclist was speeding. The solution attempts to use equations to calculate the initial speed of the motorcyclist, but there is not enough information given to accurately determine this. Additionally, the assumptions made in the solution may not be valid.
gjh
Homework Statement
A motorcyclist driving in a 60-km/h zone hits a stopped car on a level road. The motorcyclist was thrown from his bike and landed 39 m down the road. Was he speeding?
Relevant Equations
vf(x)^2=vi(x)^2+2a(x)s -> 0 = vi(x)^2+2a(39) -> a (x)=-(16.7)^2/78=-3.57 m/s^2
60-kph = (60x1000)/60x60 = 16.7 m/s
vf(x)=vi(x)+at -> 0 = 16.7 m/s-3.57m/s(t) ->t=4.6 s
s=39=vt = v(4.6) -> v= 8.4 m/s
s=16.7(4.6)=76.8 m
I assumed the cyclist was going 60-kph. Then worked backwards as shown above to calculate the deceleration in the x direction and then the time to decelerate to a stop at 39 m. Then used this time to calculate the initial speed if cyclist thrown 39 m, which was 8.4 m/s. But if he was going 16.7 m/s, he would have been thrown 76.8 m. The answer is he was speeding, so I have the logic reversed or one of the equations is wrong. Grateful for any hints.

Hmm, I am not sure I understand the problem statement. Does the problem's complete statement include anything of the following:
• kinetic friction coefficient and mass of motorcyclist
• angle of ejection of motorcyclist from his bike
• maximum height that his body reaches before "landing" down the road
• anything else additional that the problem statement says and you forgot to mention
From the data given I was unable to form a complete solution.

If you want me to judge what you did in your attempt, I find it as some sort of "spaggheti" solution because it is really complexified and I am having a hard time to understand what you did. But it might be just me because as I ve already said I haven't understand the problem statement.

gjh said:
Homework Statement:: A motorcyclist driving in a 60-km/h zone hits a stopped car on a level road. The motorcyclist was thrown from his bike and landed 39 m down the road. Was he speeding?
Relevant Equations:: vf(x)^2=vi(x)^2+2a(x)s -> 0 = vi(x)^2+2a(39) -> a (x)=-(16.7)^2/78=-3.57 m/s^2
60-kph = (60x1000)/60x60 = 16.7 m/s
vf(x)=vi(x)+at -> 0 = 16.7 m/s-3.57m/s(t) ->t=4.6 s
s=39=vt = v(4.6) -> v= 8.4 m/s
s=16.7(4.6)=76.8 m

I assumed the cyclist was going 60-kph. Then worked backwards as shown above to calculate the deceleration in the x direction and then the time to decelerate to a stop at 39 m. Then used this time to calculate the initial speed if cyclist thrown 39 m, which was 8.4 m/s. But if he was going 16.7 m/s, he would have been thrown 76.8 m. The answer is he was speeding, so I have the logic reversed or one of the equations is wrong. Grateful for any hints.
Assuming the motorcyclist was doing the speed limit is not a particularly useful start, but let's see where you went from there. You calculated the acceleration and time elapsed to come to a stop in 39m, then calculated the distance if going at a constant 60kph for that time. A bit strange since you know the speed was not constant.

A difficulty with this question is that there does not appear to be enough information. Since we do not know the initial speed, how are we to determine the acceleration? If the road were covered in ice he could have slid a long way without necessarily having been moving very fast to begin with.

But note that it says "landed" 39m further on. That implies this is not where he finished up but where he first hit the ground. Does that give you any ideas on how to proceed?

Delta2
Maybe you don't know the angle of ejection. However, you can assume that it is the angle of maximum range and that his initial speed is 60 kph and calculate where he would land under these assumptions. If the distance is less than 39 m then ##\dots##

PeroK, BvU and Delta2
gjh said:
Homework Statement:: A motorcyclist driving in a 60-km/h zone hits a stopped car on a level road. The motorcyclist was thrown from his bike and landed 39 m down the road. Was he speeding?
The question is not very good. If a motorcycle hits an obstacle, then the initial velocity of the rider is horizontal. There must be some process during the collision that changes the direction of motion, but the angle above the horizontal is unknown. Also, the change (if any) of the magnitude of velocity is unknown. The external force from the parked vehicle could propel the rider into the air - by flipping the bike up.

In other words, if the motorbike is traveling at ##60 \ km/h## when it hits the car, the rider is propelled with an unknown velocity (magnitude and direction) into the air.

You would need specific data about such collisions to draw any conclusions.

Moreover, I would say that ##39 \ m## is a long way to be propelled in the air.

PeroK said:
You would need specific data about such collisions to draw any conclusions.
Not necessarily. The velocity of the rider when thrown skywards cannot reasonably be more than the original velocity of the bike, and 45° would maximise the range, so it is possible to find a lower bound on the bike's speed.

Delta2
haruspex said:
Not necessarily. The velocity of the rider when thrown skywards cannot reasonably be more than the original velocity of the bike,
Why not? The rider may get an upwards impulse from the rotation of the bike. It's effectively a three-body collision.

PeroK said:
Why not? The rider may get an upwards impulse from the rotation of the bike. It's effectively a three-body collision.
A pole-vaulter is limited fundamentally by his forward speed. So his bar height sets a lower bound on his speed. Same for this ballistic motorcyclist and his downrange distance (ignoring air) .
So I think this data could in fact show that the cyclist must have been speeding. It cannot show definitively that he was not speeding

hutchphd said:
A pole-vaulter is limited fundamentally by his forward speed. So his bar height sets a lower bound on his speed. Same for this ballistic motorcyclist and his downrange distance (ignoring air) .
So I think this data could in fact show that the cyclist must have been speeding. It cannot show definitively that he was not speeding
There is this:

berkeman, Orodruin and hutchphd
hutchphd said:
A pole-vaulter is limited fundamentally by his forward speed. So his bar height sets a lower bound on his speed. Same for this ballistic motorcyclist and his downrange distance (ignoring air) .
So I think this data could in fact show that the cyclist must have been speeding. It cannot show definitively that he was not speeding
There is a fundamental difference here*, which is that the pole in a pole vault is significantly lighter than the vaulter and therefore adds negligible energy. This would typically not be the case on a motorcycle. There is a basic limit on the kinetic energy of the motorcyclist coming from energy conservation of the full system, but this will not place the ingoing speed as a limit.

* Technically, the pole vault max height is not actually determined solely (although mainly) by the ingoing speed either. The vaulter has plenty of opportunity to also add some energy during the actual jump and to affect the bar clerance by manipulating body position relative to the center of mass.

Ok, maybe your center of mass does pass above the bar if you are Armand Duplantis…

hutchphd and PeroK
Good fun. Thanks.
As I again think about the motorcycle collision dynamics, it becomes clear that the rider will likely receive ~zero horizontal impulse (from the bike at least) and a very situation-dependent vertical impulse depending (as @PeroK indicated) upon the rotational behavior of the bike from whence the energy comes. ∴ One can provide neither bound. So much for simple physics.
Should vaulters carry big heavy poles?...

hutchphd said:
Should vaulters carry big heavy poles?...
Probably not. That would likely slow them down too much. On the contrary, I think manufacturers make them as light as they can while maintaining appropriate stiffness and reliability.

Here's a screenshot of Duplantis' 6.18 m from Glasgow in 2020. It was a world record at that time and probably his best jump so far (both his 6.19 m and 6.20 m jumps had significantly less clearance).

hutchphd
A bit more on-topic: As demonstrated in this thread, using very rudimentary physics arguments to try to model things like traffic accidents is not very advisable as there are many many unknowns and complicating factors. This is part of why threads in the direction "Help me show that the guy that hit me in traffic was speeding!" are not appropriate for PF.

hutchphd and PeroK
kuruman said:
Maybe you don't know the angle of ejection. However, you can assume that it is the angle of maximum range and that his initial speed is 60 kph and calculate where he would land under these assumptions. If the distance is less than 39 m then ##\dots##
I think this line of reasoning works. Here's how I applied it: Assume 39 m is the max range and it was achieved at launch angle of 45 degrees. Use the range formula X=2v(0)^2 sin (theta)*cos (theta)/g, where X=39 m and theta = 45 degrees and solve for v(0) -> v(0) = 19.6 m/s, which is larger than 16.7 m/s (48 kph). So the conclusion is that the cyclist was speeding. "Feels right," but would like additonal comment.

Also, no other information was given in the problem - no launch angle, no time of flight, no max height, etc. This problem came from a chapter in Wolfson's text where he only discussed the problem of velocity and acceleration in 2 dimensions, given constant acceleration (gravity) and how it could be handled as separate problems in the x and y directions.

Again, thanks to all who found the time to comment.

hutchphd, PeroK and Delta2

Tom.G, PeroK and Delta2
gjh said:
I think this line of reasoning works. Here's how I applied it: Assume 39 m is the max range and it was achieved at launch angle of 45 degrees. Use the range formula X=2v(0)^2 sin (theta)*cos (theta)/g, where X=39 m and theta = 45 degrees and solve for v(0) -> v(0) = 19.6 m/s, which is larger than 16.7 m/s (48 kph). So the conclusion is that the cyclist was speeding. "Feels right," but would like additional comment.

Also, no other information was given in the problem - no launch angle, no time of flight, no max height, etc. This problem came from a chapter in Wolfson's text where he only discussed the problem of velocity and acceleration in 2 dimensions, given constant acceleration (gravity) and how it could be handled as separate problems in the x and y directions.

Again, thanks to all who found the time to comment.
Not to beat a dead horse ... well, maybe a little since I spent many hours a/analyzing this problem, I checked my answer by finding the range at 16.7 m/s as suggested. Using x = (v(0)^2)sin (2sin (theta)*cos (theta)/g -> x = 28 m. But we know the range was 39 m - conclusion is he was speeding. The 19.6 m/s would be the minimum speed (lower bound). One of the reasons I love Wolfson is you learn as much from his problems as you do his lectures - makes you work for it, as they say. Again, thanks to all for helping me with this problem!

berkeman
Orodruin said:
Here's a screenshot of Duplantis' 6.18 m from Glasgow in 2020. It was a world record at that time and probably his best jump so far (both his 6.19 m and 6.20 m jumps had significantly less clearance).

Update: World champion and new world record with a lot of clearance. That man is on fire!

Also surpassed Ukrainian legend pole vaulter Sergey Bubka in number of competition jumps above 6.00 m. Bubka was the first to jump 6 meters and first above 6.10 m and set 35 world records (Duplantis now has five - in meters - 6.17, 6.18, 6.19, 6.20, and 6.21, three of which were set this year.)

berkeman and Delta2
gjh said:
Homework Statement:: A motorcyclist driving in a 60-km/h zone hits a stopped car on a level road. The motorcyclist was thrown from his bike and landed 39 m down the road. Was he speeding?
Relevant Equations:: vf(x)^2=vi(x)^2+2a(x)s -> 0 = vi(x)^2+2a(39) -> a (x)=-(16.7)^2/78=-3.57 m/s^2
60-kph = (60x1000)/60x60 = 16.7 m/s
vf(x)=vi(x)+at -> 0 = 16.7 m/s-3.57m/s(t) ->t=4.6 s
s=39=vt = v(4.6) -> v= 8.4 m/s
s=16.7(4.6)=76.8 m

I assumed the cyclist was going 60-kph. Then worked backwards as shown above to calculate the deceleration in the x direction and then the time to decelerate to a stop at 39 m. Then used this time to calculate the initial speed if cyclist thrown 39 m, which was 8.4 m/s. But if he was going 16.7 m/s, he would have been thrown 76.8 m. The answer is he was speeding, so I have the logic reversed or one of the equations is wrong. Grateful for any hints.
Lower bound on speed emerges quickly from ##\vec{a} \times \vec{s} = \vec {v} \times \vec{u}\implies39g=v^2sin(180-2\theta)=v^2sin(2\theta)##

gjh said:
and solve for v(0) -> v(0) = 19.6 m/s, which is larger than 16.7 m/s (48 kph).
I get the same result, except to note that 16.7 m/s corresponds to 60 kph

Orodruin said:
A bit more on-topic: As demonstrated in this thread, using very rudimentary physics arguments to try to model things like traffic accidents is not very advisable as there are many many unknowns and complicating factors. This is part of why threads in the direction "Help me show that the guy that hit me in traffic was speeding!" are not appropriate for PF.
I agree with this point, but I think the author of the problem is just trying to add some interest to the admittedly dry ballistics. I had a professor who gave us problems where "the National Guardsman sprays a fire hose on the anti-war demonstrator..." and we had to determine the force on the demonstrator and the acceleration.

EDIT If it isn't obvious, this class was back in the 1970s

Delta2
gmax137 said:
but I think the author of the problem is just trying to add some interest to the admittedly dry ballistics
Which may be all in good faith, but it gives the false impression of actually being able to model such things.

gmax137

## What is projectile motion?

Projectile motion is the motion of an object through the air that is affected by gravity. It follows a curved path called a parabola.

## How does projectile motion apply to a motorcyclist?

A motorcyclist can experience projectile motion when they are in the air after hitting a jump or bump. Their motion is affected by gravity and they follow a parabolic path.

## What factors affect the projectile motion of a motorcyclist?

The factors that affect the projectile motion of a motorcyclist include the initial velocity, the angle at which they leave the ground, and the force of gravity.

## How can the trajectory of a motorcyclist be calculated?

The trajectory of a motorcyclist can be calculated using the equations of projectile motion, which take into account the initial velocity, angle, and acceleration due to gravity.

## What safety precautions should be taken for a motorcyclist experiencing projectile motion?

It is important for a motorcyclist to wear proper safety gear, such as a helmet and protective clothing, to minimize the risk of injury during projectile motion. They should also ensure they have proper training and experience before attempting jumps or other maneuvers that involve projectile motion.

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