Horizontal projectile, NOT GIVEN INITIAL VELOCITYby xannaxiero Tags: horizontal, initial, projectile, velocity 

#1
Feb2112, 05:43 PM

P: 10

1. The problem statement, all variables and given/known data
A cannon ball is fired horizontally off the top of a cliff that is 400 m high above the ocean. The ball hits the water 600 m away. With what speed does the cannon ball enter the water? 2. Relevant equations I've tried using d[itex]_{x}[/itex] = u[itex]_{x}[/itex]*t d[itex]_{y}[/itex] = u[itex]_{y}[/itex]*t + 1/2a[itex]_{y}[/itex]*t[itex]^{2}[/itex] and also solving for t t = [itex]\sqrt{2y/g}[/itex] 3. The attempt at a solution So I found that t = 9 seconds and that u[itex]_{x}[/itex] = 66.4 but i can't seem to figure out how to find final velocity, or speed. I know what the correct answer is (111 m/s), I just can't figure out how it was gotten. I have an exam in an hour, any help would be greatly appreciated!! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Feb2112, 06:08 PM

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P: 40,905

You found the horizontal component of velocity, now find the vertical component. Once you have both components, you can figure out the total speed.




#3
Feb2112, 06:14 PM

P: 10

u[itex]_{y}[/itex] = 22.6
so if i do square root of 66.4^2 + 22.6^2 I should get my answer but i get 70.1 instead. how do you find the final based on the x and y components? 



#4
Feb2112, 06:17 PM

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P: 40,905

Horizontal projectile, NOT GIVEN INITIAL VELOCITY 



#5
Feb2112, 06:19 PM

P: 10

t = 9 sec
so i did the following: 600 = uy + 1/2*9.8*9^2 solving for uy I got 22.6 



#6
Feb2112, 06:27 PM

P: 615





#7
Feb2112, 06:34 PM

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P: 40,905





#8
Feb2112, 06:37 PM

P: 10

ah yes. okay so that means 400 = uy*9 + 1/2 *9.8*9^2
so uy = .34 and i guess i did ux wrong, so ux is actually 600 = ux*9, giving ux = 67 but from here i still don't know how to find the final velocity 



#9
Feb2112, 06:44 PM

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P: 40,905

How does velocity change for accelerated motion? What's the definition of acceleration? 



#10
Feb2112, 06:49 PM

P: 10

velocity increases due to acceleration... a = change in velocity/change in time
so we have a change from zero to 66.4 over the course of 9 seconds meaning acceleration is 7.4 m/s ? and i know v = vi + at so i feel like i could do v= 0 + 7.4*9 giving me 66.4, which is wrong. 



#11
Feb2112, 06:52 PM

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P: 40,905




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