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Projectile Motion

R=v2sin⁡2θg=40.77 mR = \frac{v^2\sin 2\theta}{g} = 40.77\,\text{m}
Range40.8 m
Max height10.2 m
  • Bring the range down to about 25 m
  • Reach a maximum height above 15 m
  • Find the launch angle that gives the maximum range
  • Land the projectile at a range of about 50 m