A football player releases a ball with the initial conditions shown in the figure. – Free B13
A football player releases a ball with the initial conditions shown in the figure. Determine the radius of curvature of the trajectory at times 𝑡 = 0.9 sec t=0.9sec and 𝑡 = 1.9 sec t=1.9sec, where 𝑡 = 0 t=0 is the time of release from the quarterback’s hand. For each case, compute the time rate of change of the speed.
Answer
Projectile Motion: Football Throw – Detailed Physics Analysis
🔹 Parametric Equations of Motion
Given an initial speed of 79 ft/s and launch angle 330°:
x(t) = (79 cos 330°) × t
y(t) = (79 sin 330°) × t − 16t²
y(t) = (79 sin 330°) × t − 16t²
These are derived from standard projectile motion with gravitational acceleration of 32 ft/s².
🔹 Velocity Components
Taking derivatives with respect to time:
vx(t) = 79 cos 330° = 66.255 ft/s (constant)
vy(t) = 79 sin 330° − 32t
vy(t) = 79 sin 330° − 32t
🔹 Speed (Magnitude of Velocity)
v(t) = √[vx(t)² + vy(t)²]
🔹 Radius of Curvature
ρ(t) = [v(t)³] / [|v(t) × a(t)|] = [v(t)³] / [32 × |vx(t)|]
🔹 Tangential Acceleration
aₜ(t) = (vx × ax + vy × ay) / v(t)
Since ax = 0 and ay = −32, the formula simplifies accordingly.
⏱️ Evaluation at Specific Times
📍 At t = 0.9 s
- vx = 66.255 ft/s
- vy = 14.2265 ft/s
- v(0.9) = 67.77 ft/s
- ρ(0.9) = 146.8 ft
- aₜ(0.9) = −6.718 ft/s²
📍 At t = 1.9 s
- vx = 66.255 ft/s
- vy = −17.77 ft/s
- v(1.9) = 68.598 ft/s
- ρ(1.9) = 152.25 ft
- aₜ(1.9) = 8.29 ft/s²
✅ Final Summary Table
| Time (s) | Speed (ft/s) | Radius of Curvature (ft) | Tangential Acceleration (ft/s²) |
|---|---|---|---|
| 0.9 | 67.77 | 146.8 | −6.718 |
| 1.9 | 68.598 | 152.25 | 8.29 |