GSO003 Problem 4
Problem Statement
Schre's Leap: Projectile Motion of a Cat Treated as a Point Mass
Problem Statement
One day, a physics student observed his cat "Schre" make a remarkable jump.
Schre jumped from a horizontal floor, barely cleared the tip of a potted plant (an obstacle) placed in front, and landed perfectly on a favorite toy on the floor.
The student, using the physicist's bold assumption that "a cat is a point mass," set out to analyze the initial velocity of this extraordinary jump.
Define the launch point as the origin O(0,0), with the x-axis pointing horizontally to the right and the y-axis pointing vertically upward.
Schre is launched from the origin at elevation angle θ (0<θ<2π) and initial speed v0. The tip of the obstacle is at coordinates (L,H), and the toy is at (D,0).
Air resistance is negligible; the magnitude of gravitational acceleration is g.
From the conditions that Schre's trajectory passes through both (L,H) and (D,0), derive the value of v02.
Constraints
- Horizontal distance to obstacle: L=1.0 m
- Height of obstacle: H=0.50 m
- Horizontal distance to landing point: D=1.6 m
- Magnitude of gravitational acceleration: g=9.8 m/s2
Input Format
Compute v02 using the given values. The answer is an irreducible fraction BA (A,B are coprime positive integers); give the value of A+B as a positive integer.
Solution
1. Trajectory Equation
From kinematics, eliminating t:
y=xtanθ−2v02cos2θgx22. Applying the Constraints
Landing condition (D,0): since D=0,
2v02cos2θg=Dtanθ⋯②Obstacle condition (L,H): substituting ② to eliminate v02:
H=Ltanθ−L2Dtanθ=DL(D−L)tanθ tanθ=L(D−L)HD3. Substituting Numerical Values
tanθ=1.0×(1.6−1.0)0.50×1.6=0.600.80=344. Computing v02
From ②:
v02=2tanθgD(1+tan2θ)With tanθ=34: tan2θ=916, so 1+tan2θ=925.
v02=2×349.8×1.6×925=915.68×25×83=349 (m/s)2This is already an irreducible fraction: A=49, B=3.
A+B=49+3=52Answer: 52