Cannon & Cliff: Minimum Muzzle Velocity & Horizontal Flight

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing the trajectory of a cannonball shot from a cannon towards a cliff. The problem includes determining the minimum muzzle velocity required for the shell to clear a 25m tall cliff located 60m away from the cannon and calculating how far the shell lands past the cliff.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for the minimum muzzle velocity and the time of flight for the shell. There are attempts to clarify the relationship between vertical and horizontal motion and how to apply motion equations to find the landing distance.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning each other's reasoning. There is a mix of interpretations regarding the time of flight and the equations needed to solve for the landing distance. Some guidance has been offered on how to set up the equations for both vertical and horizontal motion.

Contextual Notes

There is some confusion regarding the application of the projectile motion equations, particularly in distinguishing between the time it takes to clear the cliff and the total time of flight until the shell lands. Participants are also addressing the initial conditions of the problem and the impact of the launch angle on the calculations.

steph2010
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a cannon located 60m from the base of a vertical 25m tall cliff, shoots a 15kg shall at 43 degrees above the horizontal toward the cliff (a) what must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) the ground at the top of the cliff is level, with a constant elevation of 25m above the cannon. Under the conditions of part (a) how far does the shell land past the edge of the cliff?
I got part (a) to be 32.63 m?s but i don't know where to start for part b.
 
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When the shell lands, the y distance is 0. You can find the time for this and how far out the shell lands using the same motion equations you used for part a.
 
but arent i finding the where it would land on the y axis?
 
The y-axis is height, so you need to know at what point in time the height is 0 (that is, the object is on the ground). Since the equation is a quadratic, you will have two answers. 1 will be 0 (since it starts from ground height) and the other will be the time at which it hits the ground on the other side of the parabola. Use that to find the x distance.
 
i am getting the answer to be 21.9264m is that right?
 
Not what I got, what did you get for the time?
 
t= 2.51 sec
 
Hmm.

d_f = \frac{1}{2}at^2 + v_it?
 
now i have to answers for t 2.04sec and 2.5sec
 
  • #10
Wait, 2.51 is the time that the shell clears the cliff, we want where it lands. Set the y equation equal to 0 and then plug that time (and the same velocity from part a) into the x equation.
 
  • #11
what y and x equations?
 
  • #12
Y: d_y = \frac{1}{2}at^2 + v_it
X: d_x = x_it

You're doing the same thing you did to make sure the shell cleared the cliff and find the velocity for that (you didn't know time or velocity) except now you don't know time or distance.
 
  • #13
ok so t= 5.77 sec
 
  • #14
I'm sorry I'm still not getting that and I'm 99% sure my work is correct. Why don't you show me your work.
 
  • #15
0=-25=32.64t-4.905t^2
t= -.88s t=5.77s
 
  • #16
OK, thanks.

First off, I assume the -25 is for the height of the cliff which isn't needed. We already proved that the shell would clear the cliff in part a, so start back at the cannon from ground level.

Secondly, the shell isn't shot horizontally, as your equation as it. It was shot 43º above horizontal, which affects the velocity.
 

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