Need help on this projectile motion problem

[LuckyIam]

Homework Statement

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?[/B]

Homework Equations

V_x = v_0x + a_xt
x = x₀ + v_0xt + 1/2a_xt²
v_0x = vcos(20)
v_0y = vsin(20)[/B]

The Attempt at a Solution

Every time I attempt to break down the velocity into its components I end up with a v_0y that's larger than v_0x, 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

 

[phinds]

I have not looked at your math, just at your wording about why you are concerned with what you got. Think about this: what is the initial horizontal competent of the object's velocity? What is the final horizontal component of the object's velocity? Think about the same thing for the vertical component (start and finish). Then think about this: does the height of the building have any effect on the horizontal component? How about vertical component?

 

[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...

 

[LuckyIam]

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...

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

 

[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?

 

[NTW]

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

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

 

[LuckyIam]

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?

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 that's 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

 

[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

 

[NTW]

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 that's 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

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...

 

[phinds]

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 that's 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

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.