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)

- anonymous

- jamiebookeater

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- anonymous

|dw:1343399513614:dw|

- anonymous

Show that \(R = 2H\)

- anonymous

well how do i start? anyone able to give me a hint?

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## More answers

- 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

i dont know vectors..

- 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

|dw:1343400018031:dw|

- anonymous

I am only able to form parametric equation for \(A\) \[y=-\frac{1}{2}
g*\left(\frac{x}{u}\right)^{2}+H
\]

- 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

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

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

well what do i do next? I am unable to find the trajectory equation thing for \(B\)

- experimentX

what about the velocity?

- anonymous

i am able to find the acceleration, velocity for \(A\) but not \(B\)

- experimentX

no ... i'm asking about the initial velocity of both projectiles

- anonymous

i dont know; 0?

- experimentX

what's this R anyway?

- anonymous

R is the range; the longest horizontal distance travelled by the object

- anonymous

|dw:1343402352605:dw|
abit clearer; sorry for the horrible drawing

- experimentX

trivial case v=0, R=0, H=0 ... what' the initial velocity of both projectiles .. are they same?

- anonymous

well the velocity i got for A for the x-motion is \(u\) and the y-motion is \(-gt\)

- experimentX

what about the B?

- anonymous

i dont know how to do B

- anonymous

because its projected vertically so im lost

- experimentX

i mean what is the initial velocity of B ... going up?

- anonymous

i dont know..i guess its going up..

- 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

how did you get y = (1/2)gt^2 for the second equation?

- experimentX

v^2 = y^2 + 2as => H = u^2/2g

- anonymous

okay; thanks!

- 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

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

thank you!

- experimentX

yw

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