Kinematics - Projectile Motion

  • Thread starter Insolite
  • Start date
  • #1
7
0
Question:

A cannon, located ##60.0 m## from the base of a vertical ##25.0 m## tall cliff, shoots a ##15 kg## shell at ##43.0°## above the horizontal toward the cliff. What must the minimum muzzle velocity be for the shell to clear the top of the cliff?

I have attempted a solution that I'm doubtful is correct as I have made the assumption that the shell glances the edge of the cliff at the highest point of its trajectory and I'm not certain whether this is necessary (as in it may just clear the cliff at a point in its trajectory that is before the highest point). However, without making this assumption I don't know how to proceed with the question.

Solution:

Finding the initial ##y##-component of velocity, where the max height of projectile is assumed to be the height of the cliff, ##25 m##, where ##v_{y} = 0## -
##v_{y}^{2} = v_{0y}^{2}sin^{2}{α} - 2g(25)##
##0 = v_{0y}^{2}sin^{2}{43°} - 2g(25)##
##50g = v_{0y}^{2}sin^{2}{43°}##
##v_{0y} ≈ 32.47 ms^{-1}##

Finding the initial ##x##-component of velocity, where I have reversed the set-up and calculated the time taken for the projectile to go from ##v_{0y} = 0## to ##v_{y} = 32.47 ms^{-1}## in the ##y##-axis -
##v_{y} = v_{0y} - (-g)t##
##v_{y} = gt##
##t ≈ \frac{32.47}{9.81}##
##t ≈ 3.31 s##

Finding the initial ##x##-component of velocity -
##x = (v_{0x}cosα)t##
##v_{0x} = \frac{x}{tcos43°}##
##v_{0x} = \frac{60.0}{3.31cos43°}##
##v_{0x} ≈ 24.78 ms^{-1}##

The magnitude of the initial muzzle velocity -
##v = \sqrt{v_{0x}^{2} + v_{0y}^{2}}##
##v = \sqrt{24.78^{2} + 32.47^{2}}##
##v = 40.9 ms^{-1}##

Any help or advise is appreciated.
Thanks.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,415
6,945
Finding the initial ##y##-component of velocity, where the max height of projectile is assumed to be the height of the cliff,
That's not a valid assumption. The cannon's angle is fixed. If you reduce the muzzle velocity to the point where the maximum height of the trajectory would be the height of the cliff, the point where it reaches that height might be beyond the cliff.
 
  • #3
TSny
Homework Helper
Gold Member
13,159
3,458
To get a feel for the problem, make a sketch of the setup that is roughly to scale with a projection angle 43o≈45o. Draw some trajectories for different initial speeds.

When the initial speed is such that the shell barely clears the cliff, does it look like vy = 0 when the shell is just over the cliff?

[oops, I see haruspex posted while I was constructing my post. Sorry.]
 
Last edited:

Related Threads on Kinematics - Projectile Motion

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
963
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
5
Views
11K
  • Last Post
Replies
3
Views
2K
Replies
11
Views
745
Top