https://bergstrom.org/tags/physics/Physics - Tag - Jorgen BergstromHugo -- gohugo.ioen-usJorgen BergstromSat, 25 May 2024 00:00:00 -0500https://bergstrom.org/posts/hammer_distance_with_drag/Sat, 25 May 2024 00:00:00 -0500Jorgen Bergstromhttps://bergstrom.org/posts/hammer_distance_with_drag/Physics-Based Theory and Numerical Implementation The weight of the hammer ball is 7.26 kg, the material is steel and therefore the ball radius is given by $r = (3m/(4\pi \rho_s))^{1/3}$. The drag force from the air resistance is $F_d = \rho_a v^2 C_d A$, where $A$ is the cross-sectional area. In summary, Newton’s equation in the horizontal and vertical directions can be written in incremental form: $$ \displaystyle \Delta v_x = – \frac{F_d(v)}{m} \frac{v_x}{v} \Delta t$$ $$ \displaystyle \Delta v_y = – \left[ \frac{F_d(v)}{m} \frac{v_y}{v} + g \right] \Delta t$$