December 18th, 2016, 08:24 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  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=3rq. 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=pr a=g t=t. I don't know where to go from here. Last edited by skipjack; December 18th, 2016 at 09:16 AM. 
December 18th, 2016, 09:05 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319 
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 ... $pr = v_f t  \left(\dfrac{1}{2}gt^2\right) = (gt)t + \dfrac{1}{2}gt^2 = \color{red}{\dfrac{3}{2}gt^2}$ 
December 19th, 2016, 10:31 AM  #3 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
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 10:57 AM. 
December 19th, 2016, 12:14 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,505 Thanks: 758  
December 19th, 2016, 03:07 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,640 Thanks: 1319  Quote:
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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