Physics Practice Problems Projectile Motion: Positive Launch Practice Problems Solution: A batter hits a fly ball which leaves the bat 0.95...

Solution: A batter hits a fly ball which leaves the bat 0.95 m above the ground at an angle of 61° with an initial speed of 30 m/s heading toward centerfield. Ignore air resistance.(a) How far from home plate would the ball land if not caught?(b) The ball is caught by the centerfielder, who starts running straight toward home plate at a constant speed from 105 m away, at the instant the ball is hit, and makes the catch at ground level. Find his speed.

Problem

A batter hits a fly ball which leaves the bat 0.95 m above the ground at an angle of 61° with an initial speed of 30 m/s heading toward centerfield. Ignore air resistance.

(a) How far from home plate would the ball land if not caught?

(b) The ball is caught by the centerfielder, who starts running straight toward home plate at a constant speed from 105 m away, at the instant the ball is hit, and makes the catch at ground level. Find his speed.

Solution

For this problem, we're looking for the horizontal distance the baseball travels given the magnitude and direction of the initial velocity and the height it was batted from. Then we also want to know how fast the centerfielder runs to catch the ball as it lands given his horizontal distance from the launch point.

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

In our coordinate system, 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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