2D Kinematics problem (hill projectile)

AI Thread Summary
A cannonball is launched from a cliff at 50 m/s at a 30-degree angle, with a range of 300 m. To find the height of the cliff, the vertical motion equations are applied, leading to a calculated height of 215 meters. The impact velocity before striking the ground is determined using energy conservation principles, resulting in a speed of approximately 84.36 m/s. There is discussion about the correct use of gravitational acceleration, with some preferring -9.8 m/s² for accuracy. The importance of separating known data into x and y components for solving 2D kinematics problems is emphasized.
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Homework Statement


3. A cannon ball is launched from a cliff with an initial velocity of 50 m/s 300 above the horizontal. If it’s range is 300 m.

a)What is the height of the cliff?
b) What is the impact velocity of the ball right before it strikes the ground?



Homework Equations


d=v1t+1/2at^2

d=V*t

Vav=V1+v2/2





The Attempt at a Solution



How do you find the height and impact velocity?
Horizontal: Vx=Cos30*50= 43m/s
dx= 300m

Vertical: Vy=Sin 30*50=25
a= -9.8 m/s^2

I am really stuck!
 
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You need to derive an equation for the range so that you can plug 300m into it. Try writing out the equations of motion for both the x and y directions. Then solve for the x when y=0 (i.e. when the ball hits the ground).
 
"initial velocity of 50 m/s 300 above the horizontal."

What do you mean by 300 above the horizontal?
 
sorry i meant 30 degrees
 
let the height of cliff be h
let the angle be 30 degrees
range = 50cos30 * time = 300
time = 4root3 seconds

initial vertical velocity is vsin30 = 25 m/s
h = 25 - 1/2gt^2
assuming g = 10m/s^2
= - 215 meters
height is 215 meters

impact velocity is easiest calculated using energy conservation
both the equation of motion and energy conservation will lead to the same equation and asnwer
initial energy = 1/2 m 50^2 + mg 215
final energy = 1/2mv^2
v = 84.36 m/s

I don't know if this is right
 
The time calculated seems right.

h = 25 - 1/2gt^2

I believe your equation is a little bit wrong.

Y = Vot + 1/2gt^2

Does your teacher accept g as 10 m/s^2?? I hate that...
 
no sorry i was doing this in my head but should i have acceleration as -9.8 m/s with the negative because my velocity is +

How would you do this problem?
 
I was just pointing out that your Y displacement formula has Vo multiplied by time.

Y = Vot+1/2gt^2

and yea, gravity would be negative.
 
i just feel very unsure doing this problem and when i do other problems i need to refer to my previous examples, could you give me some tips that you use to solve these easily?
Thanks
 
  • #10
Whenever you have a 2-d kinematics problem I always separate into 2 columns all the known data. In the x-direction and in the y-direction. And if you ever need to know something in specifically the x or y-axis and it seems as though you have insufficient data, then try linking the two by finding time.

That's just me though. And of course keep in my mind which equations solve for which variables: displacement, initial and final velocity, and time.
 
  • #11
thank you
 
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