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Range Projectile

Reference data and engineering information about range projectile for dynamics applications.

rangeprojectile

Overview

Engineering reference data for Range Projectile in dynamics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Momentum

p=mvp = mv

Mass × velocity.

Work

W=FdcosθW = Fd\cos\theta

Force × displacement × cos(angle).

Variables

SymbolDescriptionUnit
FFForceN
mmMasskg
aaAccelerationm/s²
vvVelocitym/s
EkE_kKinetic energyJ

Example: Throwing a Ball

A ball is thrown with an initial velocity of 25 m/s at an angle of 30 degrees to the horizontal plane.

Calculated Results

The time to reach maximum height is:

th=visin(Θ)ag=(25 m/s)sin(30)9.81 m/s2=1.27 st_h = \frac{v_i \sin(\Theta)}{a_g} = \frac{(25 \text{ m/s}) \cdot \sin(30^\circ)}{9.81 \text{ m/s}^2} = 1.27 \text{ s}

The maximum elevation of the flight is:

h=12agth2=12(9.81 m/s2)(1.27 s)2=7.91 mh = \frac{1}{2} a_g t_h^2 = \frac{1}{2} (9.81 \text{ m/s}^2)(1.27 \text{ s})^2 = 7.91 \text{ m}

The horizontal distance traveled is:

s=vi2sin(2Θ)ag=(25 m/s)2sin(2×30)9.81 m/s2=55.2 ms = \frac{v_i^2 \sin(2\Theta)}{a_g} = \frac{(25 \text{ m/s})^2 \cdot \sin(2 \times 30^\circ)}{9.81 \text{ m/s}^2} = 55.2 \text{ m}

Important Notes

  • Air resistance neglected: All calculations assume a vacuum with no air friction. In reality, drag forces significantly reduce both range and maximum height, especially at higher velocities.
  • Optimal angle: For maximum range on level ground (neglecting air resistance), the optimal launch angle is 4545^\circ, since sin(2Θ)\sin(2\Theta) reaches its maximum value of 1.
  • Symmetry: In the idealized case, the projectile's trajectory is symmetric about its highest point, meaning the time to ascend equals the time to descend.

References