Horizontal projectile, NOT GIVEN INITIAL VELOCITY

In summary, a cannon ball is fired horizontally off a 400 m cliff and hits the water 600 m away. The horizontal component of velocity is 66.4 m/s and the vertical component is 22.6 m/s. Using the kinematic equation v = vi + at, the final velocity can be found by substituting in the initial velocity (0 m/s), acceleration due to gravity (9.8 m/s^2), and time (9 seconds). This gives a total speed of 111 m/s for the cannon ball as it enters the water.
  • #1
xannaxiero
10
0

Homework Statement



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?



Homework 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]



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!
 
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  • #2
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
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
xannaxiero said:
u[itex]_{y}[/itex] = 22.6
How did you get that result?
 
  • #5
t = 9 sec

so i did the following:

600 = uy + 1/2*9.8*9^2
solving for uy I got 22.6
 
  • #6
xannaxiero said:
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
 
  • #7
xannaxiero said:
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.)
 
  • #8
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
xannaxiero said:
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?
 
  • #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
xannaxiero said:
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.
 
  • #12
What is the acceleration of a falling body?
 

1. What is a horizontal projectile?

A horizontal projectile is an object that is launched or thrown horizontally, meaning it moves parallel to the ground, without any initial vertical velocity.

2. What factors affect the motion of a horizontal projectile?

The motion of a horizontal projectile is affected by the initial launch angle, the force of gravity, and air resistance.

3. How does the initial launch angle affect the motion of a horizontal projectile?

The initial launch angle determines the direction of the projectile's motion. A higher launch angle will result in a longer horizontal distance traveled, while a lower launch angle will result in a shorter distance.

4. What is the trajectory of a horizontal projectile?

The trajectory of a horizontal projectile is a curved path, with the highest point being the peak of the projectile's motion and the lowest point being the ground level.

5. How does air resistance affect the motion of a horizontal projectile?

Air resistance, also known as drag, slows down the horizontal velocity of a projectile, causing it to travel a shorter distance than it would without air resistance.

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