
#1
Jan1013, 10:55 PM

P: 10

1. The problem statement, all variables and given/known data
A ball is launched at a 45 degree angle and lands 152.4 m away. What is the maximum height the ball will reach during its flight. I Know the answer, what I need is to find a way to derive an equation [itex]Δy = {\frac{Δxtan(ϴ)}{4}}[/itex] I have played around for hours working backwards and forwards with the equations below. Its miserable. 2. Relevant equations [itex]Δy = {\frac{gt}{2tan(ϴ)}}[/itex] All kinematic equations 3. The attempt at a solution[itex]Δy = {\frac{gt}{2tan(x)}}[/itex] was the best i got  with the help of the internet. Though with it i have no idea how to solve for t. Help:( 



#2
Jan1013, 11:52 PM

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P: 5,480

What does the maximum height depend on?
What does the horizontal displacement depend on? 



#3
Jan1113, 05:00 PM

P: 10

sorry the equation in the bottom is tangent of thetaθ, I'm not sure what you are asking, they are both dependent on time, assuming gravity is constant, apparently one is able to derive an equation and calculate max height using only the three kinematic equations.but the only known variables are angle of elevation and Δx, which is the final displacement from the origin.




#4
Jan1113, 05:40 PM

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deriving equation for minimum height in projectile motion with two give me's :) 



#5
Jan1113, 06:27 PM

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oh, Δy indicates the maximum height achieved by the ball, which is achieved at half the total time since the balls starting y position is negligible, Δx indicates the total distance covered in the x position of the coordinate system, so making the ball's starting position the origin, the coordinates for the ball at max height are ((Δx/2),Δy) and the coordinates for the ball after it lands are (Δx,0)




#6
Jan1113, 06:35 PM

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and yes, θ is launch angle above the horizontal




#7
Jan1113, 09:07 PM

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Ok, so assume some launch speed u and write the equations for x and y as functions of time. Assuming level ground, what will x be when it lands?




#8
Jan1113, 11:15 PM

P: 10

Δx = utcos(θ), Δy = ut/2sin(θ)  1/2g(t/2)^2 Δx = 152.4m




#9
Jan1113, 11:18 PM

P: 10

and (usinθ)/2g = Δy




#10
Jan1113, 11:20 PM

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Write another equation for the height at time t. 



#11
Jan1413, 05:24 PM

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ok so im going to start this problem over. ok so
in the y direction, Δy = (vsinθ)t  (1/2)gt^2; Vf = Vsinθ  gt; vf^2 = v^2sinθ^2  2gΔy in x direction, Δx = Vcosθt so im guessing i rearrange for t using the x equation, so t = Δx/v(cosθ) then i plug it into the first equation? lets see Δy = vsinθ(Δx/v(cosθ))  (1/2)g(Δx^2/v^2(cos^2θ)) if i simplify it, Δy = (Δx)tanθ  (1/2)gΔx^2/cos^2θ am i in the right direction ? 



#12
Jan1413, 07:43 PM

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#13
Jan1413, 10:10 PM

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i just have the final displacement in the x direction, i was just guessing that since the launch starts and ends on the ground that half the time wouldd be at max height, neglecting air reistance




#14
Jan1513, 12:50 AM

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