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You drop a ball from a building on the road $40\, \mathrm{m}$ below. Your friend beside you throws a second ball 1 second later at $25\,\mathrm{m/s}$, trying to hit your ball in mid-air. Assuming the aim is good, how high does it hit.

So I started by stating my variables

  • initial distance of ball one and two: $40\, \mathrm{m}$
  • final distance of ball one and two: $0$
  • Initial velocity of one: $0\,\mathrm{ m/s}$
  • Initial velocity of two: $25\,\mathrm{m/s}$

  • Final velocity of one: ? Final velocity of two: ?

  • acceleration of one and two: $9.81\,\mathrm{m/s^2}$

This is all I know. I started by using a kinematic equation on ball one: $$2ad=v_f^2-v_i^2$$

subbing what I know gives me $2(9.81)(40)=v_f^2 - 0^2\to v_f=28.01\,\mathrm{m/s}$

I used this value along with other knowns in the another equation to find time: $v_f=v_i+a t\to 28.01=0+9.81 \,t \to t= 2.855\,\mathrm{s}$

Ok, now I have the time the first ball took to fall, now I will use the exact same process for the second ball:

$$2ad=v_f^2-v_i^2 \to 2(9.81)(40)=v_f^2-25^2 \to v_f=37.54 \mathrm{m/s}$$

Now I'll plug that number into the other kinematic equation

$$v_f=v_i+a t \to 37.54=25+9.81\,t \to t=1.275\,\mathrm{ s}$$

Since this ball was thrown a second after the first, I'll add one second to the time above: $1.275+1= 2.275$.

Since this time is faster than the time for the first ball, that means they do meet. But now how do I find that out? I tried to make kinematics equations equal each other but I get nowhere since all the distances cancel out. Please help.

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    $\begingroup$ Hi Nabil.Please read the guidelines regarding homework problems by following this link;meta.physics.stackexchange.com/questions/714/… $\endgroup$ Oct 10, 2015 at 3:20
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    $\begingroup$ Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ Oct 10, 2015 at 4:52
  • $\begingroup$ Who does upvote such type of homework-questions? Really horrendous:( $\endgroup$
    – user36790
    Oct 10, 2015 at 6:54
  • $\begingroup$ Initial distance of ball 1 and 2 is NOT 40 m. 1 second after being dropped ball 1 is just 5 m ($\frac12 gt^2$) away. That should help. $\endgroup$
    – Floris
    Oct 10, 2015 at 12:20
  • $\begingroup$ Sorry I'm new to this site I didn;t know.. And Floris, could you please explain why the initial distance is not 40 for ball one? $\endgroup$ Oct 10, 2015 at 16:09

2 Answers 2

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For the second ball to reach the first ball, they must reach the same height at the same time, We should have an equation which ties this together; when the balls hit each other, an observer would have found a time $t$ on his stop watch in which$t=t_{1}$, ball two must reach that height in that time but there is a one second gap, so we have, $$t_{2}=t_{1}-1$$

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  • $\begingroup$ Please don't post complete or near-complete answers to homework-like questions. It is against the policy of this site to completely provide answer to a homework-question which doesn't ask for a specific physics concept. This is not a homework-solving site. We discourage such answers which act as complete solution to such questions. Check our homework-policy for more clarification. $\endgroup$
    – user36790
    Oct 10, 2015 at 6:53
  • $\begingroup$ @user36790 noted $\endgroup$
    – Socre
    Oct 10, 2015 at 7:32
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They both must travel the same displacement in order to collide with each other. Try using the kinematic equation $s=ut+\frac{1}{2}at^2$. Remember that as Socre said, one will have travelled the displacement in $x$ seconds, while other in $(x-1)$ seconds as it was thrown 1 second after and that the displacement of the ball 1 = the displacement of the ball 2 as they collide.

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  • $\begingroup$ So, would I just turn the equation above into ut+1/2at^2=ut+1/2at^2? $\endgroup$ Oct 10, 2015 at 16:09

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