2D Kinematics problem (hill projectile)

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Homework Help Overview

The problem involves a cannon ball launched from a cliff at an initial velocity of 50 m/s at an angle of 30 degrees above the horizontal, with a specified range of 300 m. The questions focus on determining the height of the cliff and the impact velocity of the ball just before it strikes the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss deriving equations for the range and the motion in both x and y directions. There is an attempt to clarify the angle of launch and its implications. Some participants suggest using energy conservation to find the impact velocity, while others express uncertainty about the calculations and the use of gravitational acceleration.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided guidance on separating known data into x and y components and linking them through time. There is no explicit consensus on the correctness of the calculations or methods being discussed.

Contextual Notes

Participants are questioning the use of different values for gravitational acceleration and the assumptions made regarding the equations of motion. There is also a mention of needing to refer to previous examples for similar problems.

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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 = volt + 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 = volt+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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