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I came across an interesting question today at work.

'Imagine traveling in a car. The passenger has a glass bottle in their hand. In which direction relative to the moving car should the passenger throw it to minimize the danger of it breaking on the ground?'

I have assume so far the ground in question refers to the ground outside the car. There was also no mention of the motion of the car, so I have assumed that it must be traveling at a constant velocity, v. I have come across similar questions to this in the past; from these I know that I while it appears from the passengers reference frame that the glass is stationary in their hand, from the reference frame of somebody outside the car, it actually has the same velocity v as the car since the passenger and glass are inside the car.

I am very confused on the trajectory at which the glass should be thrown to minimize its damage upon collision with the ground. I have started with idea that the horizontal and vertical components of the motion of the glass are independent.

  • If the glass was thrown forwards at a velocity $u$ in the direction of travel it would have horizontal velocity of $v+u$. Yet the acceleration and time it took to hit the ground should be equal to if i just dropped it from the same initial vertical height.
  • If the glass were thrown backwards at a velocity $u$ opposite to the direction of travel it would have horizontal velocity $\lvert v-u \rvert$ and would still take the same time to hit the ground, with the same acceleration.
  • Throwing the glass downwards seems ludicrous so my only other thought was to throw the glass straight upwards or a combination of up and horizontal (but as I said I believe the vertical and horizontal components are independent so this also confuses me).
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    $\begingroup$ If the car is moving at constant speed in a straight line then the trajectory of the bottle, no matter what you do to it, would be the same if the car was perfectly stationary. See Galilean invariance: en.wikipedia.org/wiki/Galilean_invariance $\endgroup$ – Gert Jan 26 '16 at 16:58
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As you figured out, the horizontal and the vertical part of the motion are independent.

If you throw the bottle upwards, it will go upwards for some time, then turn and fall back. When it reaches the height of your hand, it will have the same velocity as when you threw it, just the opposite direction (downwards) - so this doesnt help as well. Clearly, vertical component should be 0.

For the horizontal component, while it doesnt change the vertical component at all, you can still kind of "minimize the danger of breaking it" by throwing it at exactly $u=-v$ - leaving a horizontal component of 0. This does not change the time it needs to hit the ground, but its the lowest overall-velocity you can achieve....

For the trajectory, proceed just as you did: add up the velocity at which you throw the bottle and the velocity of the car. Then ignore the moving car and just calculate the trajectory in the ground frame.

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When you say "minimizing the danger of it breaking on the ground" I am assuming you mean you want to reduce the kinetic energy of the object when it hits the ground (Also I have assumed there is no air resistance in the problem). In order to do that the object must not have any horizontal or vertical velocity component at the moment of release wrt the reference frame of the car. (If it has an upward vertical component then it is effective as dropping the object from a greater height. And horizontal component will contribute to a greater kinetic energy when it hits the ground). In other words, it should be released from rest wrt the reference frame of the ground. Since the car is moving with velocity $v$, you must throw the object in the opposite direction with the same magnitude |$v$| so that the net velocity is zero.

This will make the object fall with the least velocity and hence least kinetic energy. Hope this answers your question.

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