A ball is thrown up , another is dropped. Do they arrive at the same t

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When a ball is thrown upwards and another is dropped, they do not arrive at the ground simultaneously. The thrown ball travels a greater distance, reaching a maximum height before descending, while the dropped ball falls directly to the ground. The time taken for each ball can be calculated using kinematic equations, considering their respective distances and initial velocities. The thrown ball effectively starts its descent from a higher point, leading to a longer total travel time. Conducting an experiment with two balls can illustrate this concept practically.
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Homework Statement



A ball is thrown up , another is dropped. Do they arrive at the same time?


Homework Equations





The Attempt at a Solution



I think they won't because the ball which is thrown up has a greater distance to travel, but I am not sure, can someone explain please?
 
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Arrive where?
 
the ground of course..
 
If I get the question right, you should do the following;

Assume that the current height of the balls is x. The first one is thrown up whereas the second is dropped. Distance the second ball travels can be found by using the following equation ; x=x0 + V0.t + 1/2at²

Now, let's say the first ball will go h higher hence the total distance travels is 2h+x. Now, you can do the same above equation and can find the total time.

Hope, I'm not mistaken.
 
Hey, what the, just get two balls and perform the experiment.
 
There is an easy way to look at the problem. The ball that is thrown upwards will eventually come to a halt at some height H. So in effect you have two balls that are being dropped from different heights...and one has a head start.
 

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