Ch 03: 2D Motion (Projectile Motion)WorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Solution: 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 hit the top edge of the cliff?

Solution: 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 hit

Problem

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 hit the top edge of the cliff?

Solution

This problem is asking us to find the minimum initial velocity of the cannonball given the direction of the initial velocity and the horizontal and vertical distance it has to travel.

For projectile motion problems in general, we'll follow these steps to solve:

  1. Identify the target variable and known variables for each direction—remember that only 3 of the 5 variablesx or Δy, v0, vf, a, and t) are needed for each direction. Also, it always helps to sketch out the problem and label all your known information!
  2. Choose a UAM equation—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.
  3. Solve the equation for the target (or intermediate) variable, then substitute known values and calculate the answer.

The four UAM (kinematics) equations are:

 vf = v0 +atx= vf+v02tx= v0t+12at2 vf2= v02 +2ax

We define our coordinate system so that the +y-axis is pointing upwards and the +x-direction is horizontal along the launch direction. That means ay = −g, and ax = 0 (because the only acceleration acting on a projectile once it's launched is gravity.)

For projectiles with a positive launch angle, we also need to know how to decompose a velocity vector into its x- and y-components:

v0x=|v0| cos θv0y=|v0| sin θ

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