1
$\begingroup$

During a conference I have listened that an employee of an armed force, he has said that when a projectile fired from a gun upwards, it is very dangerous because when the projectile descends downwards it acquires a higher speed than the initial instant that is when the projectile was fired upwards.

In my opinion all this is false if we do not assume in the meantime that there is air resistance (friction) and from what height it is fired. Obviously if the air resistance is zero the initial velocity of the projectile $v_0$ will be equal to the final velocity of the same when it reaches the ground.

For example if $v_0=2$ m/s is the velocity of the projectile when it is fired upwards, the speed of descent $v_d$, for the opinion of this officer, is such that: $$v_0<v_d$$

I simply don't agree because if I was facing a balcony and someone shoots me from the ground at an altitude $h>h_0=0$ it hurts less if the same projectile falls on my head at the time of the descent?

$\endgroup$
  • $\begingroup$ What's not clear? Why is there a closure? I have explained what has been asked of me and what I have heard. $\endgroup$ – Sebastiano Mar 25 '19 at 22:36
  • 1
    $\begingroup$ The first sentence is not really clear. "impact with my body it hurts less"? You should try to formulate this question with some more algebra. $\endgroup$ – Roy Simpson Aug 25 '19 at 20:26
  • $\begingroup$ @RoySimpson Hi, can you see if the my edit is more clelar, please? $\endgroup$ – Sebastiano Aug 25 '19 at 20:34
  • 1
    $\begingroup$ It is clearer English, but the question is still mathematically vague: I think that this is why the Answerer below thought you were asking about running away from the bullet. The initial velocity is $v0$, the velocity an upward shot reaches height $h$ is $v1$ and the velocity of a returning trajectory reaching height $h$ is $v2$. I think that you are asking about the relationship between $v0$, $v1$ and $v2$. $\endgroup$ – Roy Simpson Aug 25 '19 at 20:48
  • $\begingroup$ @RoySimpson Look the new edit, please. Thank you very much. $\endgroup$ – Sebastiano Aug 25 '19 at 20:55
1
$\begingroup$

Can you please add some context on what are you trying to compare exactly? I might be oversimplifying the problem, but the way I see it, I would agree with the employee:

  • If the maximum height of the projectile fired upwards is equal to the height the other projectile is dropped, then we would have around the double of time to run away (thus less dangerous in that sense);

  • If the time to run away is equal in both cases, it means that the maximum height is smaller in the former case, thus the impact velocity if lower, being again the first case, less dangerous.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.