Relative position and velocity question (Two dimensions)

[kappakeepo]

Homework Statement

A freight train is moving at a constant speed of 10 m/s. A man standing on a flatcar throws a ball into the air and catches it as it falls. Relative to the flatcar, the initial vellocity of the ball is 15 m/s straight up.
a. What are the magnitude and direction of the initial velocity of the ball as seen by a second man standing on the ground next to the track?
b. How much time is the ball in the air according to the man on the train? According to the man on the ground? Mathematical prove your two answers.
c. What horizontal distance has the ball traveleed by the time it is caught according to the man on the train? According to the man on the ground?
d. What is the minimum speed of the ball during its flight according to the man on the train? According to the man on the ground?
e. What is the acceleration of the ball according to the man on the train? According to the man on the ground?

Homework Equations

Flatcar = Train I think?
C=Car
B=Ball
G=Ground(the second man)

r[B,G] = r[B,C] + r[C,G]

The Attempt at a Solution

r[B,C] = 0 i' + (Vi,y[B,C] *t -0.5g*t^2) j'
r[C,G] = (Vi,x[C,G] *t) i' + 0 j'

for v equations just dr/dt it.

Thus,

r[B,G] = 0 i' + (Vi,y[B,C] *t -0.5g*t^2) j' + (Vi,x[C,G] *t) i' + 0 j'
= (Vi,x[C,G] *t) i' + (Vi,y[B,C] *t -0.5g*t^2) j'

I'm just putting my answer for a) here so maybe if you guys want to verify if my solution is accurate.

V[B,G] = V[B,C] + V[C,G]
= (Vi,x[C,G]) i' + (Vi,y[B,C] -gt) j'
= (10) i' + (15-10t) j'
Since it asked for initial velocity, t=0 I guess?
= 10i' + 15j'I'm just going straight to the point. For b),

Now as to my understanding, the ball is on air as long as r>0. Thus solving for r=0 for r[B,C], I arrived at t=3s. Now my problem is r[B,G]. I'm pretty sure t=3s as well for it, but problem is solving it mathematically as the question wanted. All I did was stated since ball is in the air only when j' > 0, then I ignored(deleted) the i' component from the equation of r[B,G], then solving it at arrive at t=3s. I wonder if there is any other way to solve it?

Oh, and btw, for e), I got dV[B,C]/dt = a[B,C] = -g j' and dV[B,G]/dt=a[B,G] = -g j'. I think that's the answer right because I remember that acceleration is the same to all observers.

 

[Simon Bridge]

V[B,G] = V[B,C] + V[C,G]
= (Vi,x[C,G]) i' + (Vi,y[B,C] -gt) j'

... OK - I think I've sorted out the notation.
Initially it looked like you were multiplying a velocity Vi by a position x[C,G] etc.

Try this:

Let $\vec{u}$ indicate an initial velocity:
$$\vec u_{bg} = \vec u_{bc} + \vec u_{cg}\\ \vec u_{bc} = 15\hat\jmath \\ \vec u_{cg} = 10\hat\imath\\ \implies \vec u_{bg} = 10\hat\imath + 15\hat\jmath$$ ... see hot that is clearer?
There is no need to subscript the i x and y because you can define the variables at the start.
Using u for initial speed leaves you "v" available for speeds at a future time.
i.e. $$\vec v_{bg}(t) = u_{cg}t\hat\imath + (u_{bc}t-gt^2)\hat\jmath$$

For part (b) - you had already worked out the position-time equation for each POV.
It was a matter of solving each one for time. The other approach would be graphical - but it ends up with much the same actual algebra.

For (e) well done - you can also use your knowledge of other areas of physics to reality-check.
i.e. you can ask yourself what forces are on the ball in each case... you know an equation that relates force to acceleration.

 

[kappakeepo]

... OK - I think I've sorted out the notation.
Initially it looked like you were multiplying a velocity Vi by a position x[C,G] etc.

Try this:

Let $\vec{u}$ indicate an initial velocity:
$$\vec u_{bg} = \vec u_{bc} + \vec u_{cg}\\ \vec u_{bc} = 15\hat\jmath \\ \vec u_{cg} = 10\hat\imath\\ \implies \vec u_{bg} = 10\hat\imath + 15\hat\jmath$$ ... see hot that is clearer?
There is no need to subscript the i x and y because you can define the quantities later.

For part (b) - you had already worked out the position-time equation for each POV.
It was a matter of solving each one for time. The other approach would be graphical - but it ends up with much the same actual algebra.

For (e) well done - you can also use your knowledge of other areas of physics to reality-check.
i.e. you can ask yourself what forces are on the ball in each case... you know an equation that relates force to acceleration.

Thanks for your reply.

$$\vec r_{bg} = (\vec u_{cg}t)\hat\imath + (\vec u_{bc}t - 0.5gt^2)\hat\jmath$$

My only question is how do I solve the part "How much time is the ball in the air according to the man on the ground?"

I can't let $$\vec r_{bg} = 0$$ because that would be wrong obviously because here my ground is the person. And since if that is equals to 0 it would imply they are both touching/colliding each other. I only want to find how long is the ball on air i.e. the $$\hat\jmath = 0$$

I stated in my solutions that my equation ignored the unit $$\hat\imath \implies \vec r_{bg} = 0 + (\vec u_{bc}t - 0.5gt^2)\hat\jmath$$ and then solving it but I have no idea if that's acceptable. It looks wrong to me anyway still lol. Is there any better way to solve it? =)

 

[Simon Bridge]

The initial position of the ball is: $r_{bg}(t=0)=0\hat\imath + 0\hat\jmath$
The final position of the ball, wrt ground, would be $\vec r_{bg}(T) = x\hat\imath + 0\hat\jmath$
[edit]
... now you need to use your knowledge of physics: how would you normally go about finding the horizontal distance x traveled in this context?

 

[siddharth23]

A hint for finding the horizintal distance. If you're on the frieght train, the relative velocity of the ball to the train is 15 m/s vertical. There is no horizontal component. Only if you can see the train move, you can calculate the horizontal distance traveled by the ball.