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Omniscience
Group Title
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)
 2 years ago
 2 years ago
Omniscience Group Title
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)
 2 years ago
 2 years ago

This Question is Closed

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
dw:1343399513614:dw
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
Show that \(R = 2H\)
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
well how do i start? anyone able to give me a hint?
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
i dont know vectors..
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.1
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) :)
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.1
dw:1343400018031:dw
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
I am only able to form parametric equation for \(A\) \[y=\frac{1}{2} g*\left(\frac{x}{u}\right)^{2}+H \]
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
well what do i do next? I am unable to find the trajectory equation thing for \(B\)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
what about the velocity?
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
i am able to find the acceleration, velocity for \(A\) but not \(B\)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
no ... i'm asking about the initial velocity of both projectiles
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
i dont know; 0?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
what's this R anyway?
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
R is the range; the longest horizontal distance travelled by the object
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
dw:1343402352605:dw abit clearer; sorry for the horrible drawing
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
trivial case v=0, R=0, H=0 ... what' the initial velocity of both projectiles .. are they same?
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
well the velocity i got for A for the xmotion is \(u\) and the ymotion is \(gt\)
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
what about the B?
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
i dont know how to do B
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
because its projected vertically so im lost
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
i mean what is the initial velocity of B ... going up?
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
i dont know..i guess its going up..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
how did you get y = (1/2)gt^2 for the second equation?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
v^2 = y^2 + 2as => H = u^2/2g
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
okay; thanks!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

Omniscience Group TitleBest ResponseYou've already chosen the best response.0
thank you!
 2 years ago
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