NTNUJAVA Virtual Physics Laboratory
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"Progress, therfore, is not and accident,¡K" ..."Herbert Spencer(1820-1903, British philosopher)"
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Author Topic: Checking your unstanding about projectile motion  (Read 13213 times)
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Fu-Kwun Hwang
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on: February 23, 2007, 09:23:39 pm »


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The velocity components of blue curve are Vx,Vy , the velocity components of red curve are Vx,Vy.
What is the relation between Vx,Vy and Vx,Vy

(A) Vx= 2Vx, Vy=2Vy
(B) Vx=Vx, Vy=2Vy
(C) Vx=√2 Vx, Vy= √2 Vy
(D) Vx=Vx, Vy=√2 Vy

After you have your answer ready, Click the mouse inside the applet region and hit "h" key to see how the curve were set up. You can modified the equations for any motion you want to study. Just enter your forumla into the textfield. (hit "h" will toggle the display mode) Enjoy it! Smiley
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Reply #1 on: September 22, 2008, 08:40:36 am » posted from:SINGAPORE,SINGAPORE,SINGAPORE

Click the following image to view more content:

rememer to hit "h" will toggle the display mode, to see the equations

answer is (C) Vx=√2 Vx, Vy= √2 Vy

by solving by differentiation

since y1 = 4.t - 2.t^2
dy1/dt = 4 - 4.t

y2 = Math.sqrt(2).4.t - 2.t^2
dy2/dt = Math.sqrt(2).4 - 4.t

realizing when t = 0 gives the same answer as t = end point for vy1 and vy2

vy1 = dy1/dt = 4 - 4.t = 4 - 4.0 = 4
vy2 = dy2/dt = Math.sqrt(2).4 - 4.t =Math.sqrt(2).4 - 4.t = Math.sqrt(2).4 .

therefore vy2 = Math.sqrt(2).vy1


x1 = t
vx1 = dx1/dt = 1 = vx1 (constant so at end point is the same)

x2 = Math.sqrt(2).t
vx2 = dx2/dt = Math.sqrt(2)

vx2 = Math.sqrt(2).vx1

Fu-Kwun Hwang
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Reply #2 on: September 22, 2008, 01:29:45 pm » posted from:Taipei,T'ai-pei,Taiwan

I hope user do not need to use the above equations to solve the problem. Instead, Ihope user can use physics reasoning to solve it. Click the image to find out what I mean:

1. The ratio of horizontal distance between red curve and blue curve is 2. Because dx= vx*dt,  vx*dt = 2 vx*dt
2. The velocity of vy will be zero when it reach the top point. Because vy=vy0 -gt
So the time to reach the top (=dt/2) will doubled if the initial velocity is also doubled.
The maximum height h is proportional to t2: if you double velocity in vertical direction, the time to reach the top will be doubled, and the distance is velocity * time, so distance will be 4 times (2 dt * 2 vy=4*vy*dt)

Since dt = dt
So the only possible solution is Vx=√2 Vx, Vy= √2 Vy

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