🤓 Based on our data, we think this question is relevant for Professor Hatch's class at UMASS.

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?

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 a projectile 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+v02)tx= 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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