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December 18th, 2016, 09:24 AM   #1
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projectile

A particle is projected vertically upwards from a point P. At the same instant, a second particle is let fall from rest vertically at q. q is directly above p. The 2 particles collide at a point r after t seconds. When the 2 particles collide, they are travelling at equal speeds. Prove that |pr|=3|rq|.

I am trying to solve this with uvast equations. For the first particle, I have v=v s=r a=-g t=t; for the second particle, I have v=v s=p-r a=g t=t. I don't know where to go from here.

Last edited by skipjack; December 18th, 2016 at 10:16 AM.
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December 18th, 2016, 10:05 AM   #2
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for the dropped particle ...

$v_0 = 0 \implies |qr| = \color{red}{\dfrac{1}{2}gt^2}$

and its speed at collision is $v_f = gt$


for the particle shot upward ...

$|pr| = v_f t - \left(-\dfrac{1}{2}gt^2\right) = (gt)t + \dfrac{1}{2}gt^2 = \color{red}{\dfrac{3}{2}gt^2}$
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December 19th, 2016, 11:31 AM   #3
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Are you not told they are only travelling at equal speeds with they collide so how do you know that the initial speed of the particle projected upwards is vf=gt? Is the formula not s=ut+.5at² where u stands for initial speed?

Last edited by skipjack; December 19th, 2016 at 11:57 AM.
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December 19th, 2016, 01:14 PM   #4
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December 19th, 2016, 04:07 PM   #5
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Quote:
Originally Posted by markosheehan View Post
Are you not told they are only travelling at equal speeds with they collide so how do you know that the initial speed of the particle projected upwards is vf=gt? Is the formula not s=ut+.5at² where u stands for initial speed?
yes, it is ... I prefer to write it as $s = v_0 \cdot t + \dfrac{1}{2}at^2$

also, you are familiar with $v_f = v_0 + at$, correct?

solve the last equation for $v_0$ ...

$v_0 = v_f - at$

ok so far?

substitute $(v_f - at)$ for $v_0$ in the first displacement equation ...

$s = (v_f - at) \cdot t + \dfrac{1}{2}at^2$

$s = v_f \cdot t - at^2 + \dfrac{1}{2}at^2$

combine like terms ...

$s = v_f \cdot t - \dfrac{1}{2}at^2$

for projectile motion, $a = -g$ ...

$s = v_f \cdot t - \dfrac{1}{2}(-g)t^2$

$s = v_f \cdot t + \dfrac{1}{2}gt^2$

speed of the dropped object = speed of object moving upward = $gt$ ...

$s = gt \cdot t + \dfrac{1}{2}gt^2$

$s = gt^2 + \dfrac{1}{2}gt^2 = \dfrac{3}{2}gt^2$
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