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When we throw or drop an object while moving, do we impart it our acceleration or velocity or both?

Say i throw a ball at 5 m/s (relative to me) while at say, 30° to my direction of motion, then will the ball's speed in the ground frame be the vector addition of my speed and it's speed or will it be just 5 m/s?

Consider another case in which i drop an object outside from my car window, will it continue accelerating with my cars acceleration of will it just possess the instantaneous velocity of my car at that instant?

In any case, what causes it to retain any information about it's earlier motion? Why doesn't it just drop?

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  • $\begingroup$ If you drop a coke can out of your car window when doing 70mph, does the coke can keep up? $\endgroup$ – Solar Mike Apr 26 at 11:00
  • $\begingroup$ @SolarMike In space, without air resistance, it would. $\endgroup$ – WBT Apr 26 at 13:36
  • $\begingroup$ @WBT So the OP is in space - must have missed that ... $\endgroup$ – Solar Mike Apr 26 at 13:53
  • $\begingroup$ @SolarMike Not necessarily, but OP does appear to be in a universe governed by the same physical laws. Your first comment is misleading because the answer to your comment-question is strongly affected by another force (air resistance) and it's not clear if OP wants to model that at this point or not. If not, it can be helpful to think about the physics of what would happen in a setting lacking that complicating factor. $\endgroup$ – WBT Apr 26 at 13:55
  • $\begingroup$ @WBT so opening a car window into a vacuum... Or do you have a weird definition of "car"... $\endgroup$ – Solar Mike Apr 26 at 13:58
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When we throw or drop an object while moving, do we impart it our acceleration or velocity or both?

Once you release the object the only force (neglecting air friction) on the body is gravity (force downward). Its horizontal motion will be its horizontal velocity at the release point. There is no horizontal acceleration after release.

Say i throw a ball at 5 m/s(relative to me) while at say, 30° to my direction of motion, then will the ball's speed in the ground frame be the vector addition of my speed and it's speed or will it be just 5 m/s?

Assuming your motion is horizontal, and that you are not moving near the speed of light, yes it will be the vector addition.

Consider another case in which i drop an object outside from my car window, will it continue accelerating with my cars acceleration of will it just possess the instantaneous velocity of my car at that instant?

It will have no horizontal acceleration once released since there is no horizontal force acting on it (again, neglecting air friction). So when you release it it will have your horizontal velocity at the time of release and initially no vertical velocity unless thrown downward. It will have a downward acceleration of $g$.

In any case, what causes it to retain any information about it's earlier motion? Why doesn't it just drop?

It's called Newton's first law, a body in motion will continue in motion and a body at rest will remain at rest unless acted upon by an external force.

Hope this helps.

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  • $\begingroup$ Ah thanks man. That last statement really cleared up everything. $\endgroup$ – user226375 Apr 26 at 13:37
  • $\begingroup$ @user226375 You are very welcome $\endgroup$ – Bob D Apr 26 at 13:48
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When we throw or drop an object while moving, do we impart it our acceleration or velocity or both?

and

Consider another case in which i drop an object outside from my car window, will it continue accelerating with my cars acceleration of will it just possess the instantaneous velocity of my car at that instant?

Velocity only. It is the only information that is present in that object you throw. Surely, for the object to accelerate as well, you need a force acting on that object. What drives the car is the friction between the road and the tyres, but then that does not act on the object you throw out.

Say i throw a ball at 5 m/s(relative to me) while at say, 30° to my direction of motion, then will the ball's speed in the ground frame be the vector addition of my speed and it's speed or will it be just 5 m/s?

For the observer on the ground, the ball's horizontal speed will be your speed $v$ added together with the $5\cos(30^0)m/s$ of the ball with respect to your view - i.e. $v+5\cos(30^0) m/s$.

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  • $\begingroup$ Say I am running at constant velocity and throw a ball upwards, will it fall back into my hands ? $\endgroup$ – user226375 Apr 26 at 12:02
  • $\begingroup$ Also, when I throw something out of my car, why can't it keep up with me even for a little time? Is it the drag? $\endgroup$ – user226375 Apr 26 at 12:04
  • $\begingroup$ To your first question, yes. For the second question, the force of gravity acts as well, causing a downward acceleration of $9.81m/s^2$ which is very big since initially the ball had no velocity. $\endgroup$ – KV18 Apr 26 at 14:17

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