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).
Best Answer
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.