# Need help on this projectile motion problem

1. Oct 22, 2014

### LuckyIam

1. The problem statement, all variables and given/known data
A cannonball is fired from the top of a sheer cliff that towers over the valley fllor below. The angle of the initial velocity is 20 degrees above the horizontal. The cannonball remains in the air for 3.50 seconds before hitting its target on the valley floor, a distance of 80.0 m from the base of the cliff.

a) find the x- and y- components of the initial velocity of the cannonball
b) how high is the cliff?

2. Relevant equations
Vx = v0x + axt
x = x0 + v0xt + 1/2axt2
v0x = vcos(20)
v0y = vsin(20)

3. The attempt at a solution
Every time I attempt to break down the velocity into its components I end up with a v0y thats larger than v0x, which doesn't seem right at this angle.

X component:

x/t = Vi ---------> 80m/3.5s = Vicos(20) --------> 22.857/cos(20) = Vi = 24.32 m/s

Y component: for 't' at 1.75

Vf - (g)(t) = Vi -------> 0 - (-9.8)(1.75s) = Visin(20) ----------->17.15/sin(20) = Vi = 50.14 m/s

Any help would be appreciated

2. Oct 22, 2014

### phinds

3. Oct 22, 2014

### NTW

A hint...

You know that the cannonball has been flying for 3,5 seconds. And, in that time, it has covered -horizontally- a distance of 80 m.

The horizontal component of the cannonball's movement is uniform, where v = e/t. Now, you can find that horizontal component of the velocity; from that component, knowing the angle, the initial velocity of the cannonball and its vertical component can be calculated, and...

4. Oct 22, 2014

### LuckyIam

I've found the components..I want to know if they're right. It seems like the initial Vy component shouldn't be a greater value than the Vx component

5. Oct 22, 2014

### phinds

Right. You might want to rethink that 1.75 That is, why do you even think 1.75 has any particular meaning in this problem?

6. Oct 22, 2014

### NTW

Right... The vertical component in much smaller than the horizontal component. Not a surprise, since the angle with the horizontal is quite small...

7. Oct 22, 2014

### LuckyIam

Because I was trying to calculate the velocity of Vy initial from its max height point which is at half of total time in the air. I've since learned that thats not the correct of way doing it, but I'm still a little confused about how to relate each component to 20 degrees above the X axis

8. Oct 22, 2014

### Khanh

I think you should write 2 equations : x=x0+v0xt+1/2at^2 and y= y0+v0yt-1/2gt^2. Because this is projectile motion, along x-axis acceleration is 0 and you can substitute t=3.5 and x=80 into x equation=> v0x
from this you can find v0 by v0x=v0cos20=> find v0y

9. Oct 23, 2014

### NTW

Don't get confused, and try not to get lost with the equations...

Think...!

You have the time of the cannonball's flight, 3,5 s, and you know the horizontal distance covered (80 m) Remember that that horizontal motion is uniform, and hence the equation applicable is v = e/t. Thus, v = 80/3,5 => v= 23 m/s. That's the horizontal component.

Now you should draw the velocity vector, with its horizontal and vertical components. You already know the angle with the horizontal (20º). Now go ahead and determine the vertical component, It's not difficult. Just try...

10. Oct 23, 2014

### phinds

You've learned that this is not the correct way of doing it, or you've learned that 1.75 is NOT the half-way point, which is what I was getting at? 1.75 has absolutely no significance to this problem. That's what I was trying to help you understand.