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Horizontal projectile, NOT GIVEN INITIAL VELOCITY

by xannaxiero
Tags: horizontal, initial, projectile, velocity
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xannaxiero
#1
Feb21-12, 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
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Doc Al
#2
Feb21-12, 06:08 PM
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You found the horizontal component of velocity, now find the vertical component. Once you have both components, you can figure out the total speed.
xannaxiero
#3
Feb21-12, 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?

Doc Al
#4
Feb21-12, 06:17 PM
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Horizontal projectile, NOT GIVEN INITIAL VELOCITY

Quote Quote by xannaxiero View Post
u[itex]_{y}[/itex] = 22.6
How did you get that result?
xannaxiero
#5
Feb21-12, 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
SHISHKABOB
#6
Feb21-12, 06:27 PM
P: 614
Quote Quote by xannaxiero View Post
t = 9 sec

so i did the following:

600 = uy + 1/2*9.8*9^2
solving for uy I got 22.6
you want to use the displacement in the y-axis, not the x-axis
Doc Al
#7
Feb21-12, 06:34 PM
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Quote Quote by xannaxiero View Post
t = 9 sec

so i did the following:

600 = uy + 1/2*9.8*9^2
solving for uy I got 22.6
For one thing, you are dealing with the vertical motion, not the horizontal. For another, you'll need a different kinematic equation--one relating velocity and time. (Using the equation you chose will trivially give you a zero initial velocity--which you already should know.)
xannaxiero
#8
Feb21-12, 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
Doc Al
#9
Feb21-12, 06:44 PM
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Quote Quote by xannaxiero View Post
ah yes. okay so that means 400 = uy*9 + 1/2 *9.8*9^2
so uy = .34
No. Done more accurately, you'll get uy = 0. (9 seconds is just an approximate answer.) Realize that that equation gives the initial velocity, which is not what you want. You should know that the initial vertical velocity is zero. (You used that fact to find the time.)
and i guess i did ux wrong, so ux is actually 600 = ux*9, giving ux = 67
No, you were right the first time. (How did you find the time?)

How does velocity change for accelerated motion? What's the definition of acceleration?
xannaxiero
#10
Feb21-12, 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.
Doc Al
#11
Feb21-12, 06:52 PM
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Quote Quote by xannaxiero View Post
velocity increases due to acceleration... a = change in velocity/change in time
Good. Now apply that to the vertical motion.

so we have a change from zero to 66.4 over the course of 9 seconds
meaning acceleration is 7.4 m/s ?
No, the horizontal motion is not accelerated.

and i know v = vi + at
That's the equation you need. Apply it to the vertical motion.
Integral
#12
Feb21-12, 06:54 PM
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What is the acceleration of a falling body?


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