anonymous
  • anonymous
Projectile Motion: Two objects are projected with the same speed simulataneously. The object A is projected horizontally from the height H and B vertically from the ground at mid range of A. The objects later collide at some height above the ground. (will draw diagramn below)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1343399513614:dw|
anonymous
  • anonymous
Show that \(R = 2H\)
anonymous
  • anonymous
well how do i start? anyone able to give me a hint?

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anonymous
  • anonymous
Where to start? With a FBD of course! (free body diagram) ^_^ That's the easy step Show the forces as vectors acting on each object
anonymous
  • anonymous
i dont know vectors..
anonymous
  • anonymous
The goal of that first step is so you can see what is acting on a frame of reference What forces are acting on A? What forces are acting on B? Can you make two parametric equations for these? (you can) :-)
anonymous
  • anonymous
|dw:1343400018031:dw|
anonymous
  • anonymous
I am only able to form parametric equation for \(A\) \[y=-\frac{1}{2} g*\left(\frac{x}{u}\right)^{2}+H \]
anonymous
  • anonymous
The "down" is gravity acting on each object's mass, in other word the weight. The up arrow and right arrow are the initial directions of motion
anonymous
  • anonymous
Consider the horizontal and vertical motions of each: \(\sum \text{Forces}_y = \ ?\) , \(\sum \text{Forces}_x = \ ?\) Which wins out? Gravity right? What's the motion of object A for example? Upside down parabola right? Why?
anonymous
  • anonymous
This is actually something you can solve in your head by just thinking about it visually. The motion of object A is going to be (ignoring air friction, which if you wanted to be sneaky you could use to disprove the question how it was asked as false btw), moving down at an accelerating rate, t\(^2\); whereas the horizontal motion is strictly a linear rate constant, t. What you're finding here is nothing more than the intersection of two parametric curves. The parameter here is time, and there's an x & y equation for each object. The part that makes this easier is that equation for the x (horizontal) direction for object B is literally \(F_{2,x}(t) : x_2 = 0\). Am I making sense?
anonymous
  • anonymous
well what do i do next? I am unable to find the trajectory equation thing for \(B\)
experimentX
  • experimentX
what about the velocity?
anonymous
  • anonymous
i am able to find the acceleration, velocity for \(A\) but not \(B\)
experimentX
  • experimentX
no ... i'm asking about the initial velocity of both projectiles
anonymous
  • anonymous
i dont know; 0?
experimentX
  • experimentX
what's this R anyway?
anonymous
  • anonymous
R is the range; the longest horizontal distance travelled by the object
anonymous
  • anonymous
|dw:1343402352605:dw| abit clearer; sorry for the horrible drawing
experimentX
  • experimentX
trivial case v=0, R=0, H=0 ... what' the initial velocity of both projectiles .. are they same?
anonymous
  • anonymous
well the velocity i got for A for the x-motion is \(u\) and the y-motion is \(-gt\)
experimentX
  • experimentX
what about the B?
anonymous
  • anonymous
i dont know how to do B
anonymous
  • anonymous
because its projected vertically so im lost
experimentX
  • experimentX
i mean what is the initial velocity of B ... going up?
anonymous
  • anonymous
i dont know..i guess its going up..
experimentX
  • experimentX
Eqn for 1 x = ut y = H -1/2 gt^2 Eqn for 2 y = 1/2 gt^2 x = R/2 solve those two R/2 = ut t = R/2u y = H - 1/2 g (R/2u)^2 = H - gR^2/8u^2 --- 1 y = gR^2/8u^2 --- 2 H - gR^2/8u^2 = gR^2/8u^2 H = 1/4 gR^2/u^2
anonymous
  • anonymous
how did you get y = (1/2)gt^2 for the second equation?
experimentX
  • experimentX
v^2 = y^2 + 2as => H = u^2/2g
anonymous
  • anonymous
okay; thanks!
experimentX
  • experimentX
H = 1/2R^2/H => 2H = R always make sure you ask question correctly. the equation of motion of projectile is given by \[ x = u_xt \\ y = v_y + 1/2 gt^2 \] in your case horizontal component is absent in one while vertical is absent in another.
experimentX
  • experimentX
and the horizontal v of one was u while vertical velocity of another was u ... more accurately, the parametric equation of motion is given by \[ x = x_0 + ... \\ y = y_0 + ... \] if they collide means they you have to equate x's and y's of 1 and 2 ... you should be able to solve them
anonymous
  • anonymous
thank you!
experimentX
  • experimentX
yw

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