Unit 2.7 – Projectile Motion

• Definition: Projectile motion is the motion of an object that is projected into the air and moves under the influence of gravity alone (neglecting air resistance).
• Key Characteristics:
  • The motion occurs in two dimensions: horizontal ($x$) and vertical ($y$).
  • Horizontal motion is uniform (constant velocity).
  • Vertical motion is uniformly accelerated due to gravity ($g=9.8{\text{m/s}}^2$).

2. Key Concepts in Projectile Motion

(a) Trajectory:

• Definition: The curved path followed by a projectile during its motion.
• Shape: Parabolic (if air resistance is neglected).

(b) Components of Velocity:

• Horizontal Component ($v_x$):
  • Remains constant throughout the motion (no horizontal acceleration).
  • $v_x=v_0\cos \theta$, where $v_0$ is the initial velocity and $\theta$ is the angle of projection.
• Vertical Component ($v_y$):
  • Changes due to the acceleration caused by gravity.
  • $v_y=v_0\sin \theta -gt$, where $t$ is time.

(c) Time of Flight ($T$):

• Definition: Total time the projectile remains in the air.
• Formula:
  $$T=\frac{2v_0\sin \theta }{g}$$

(d) Maximum Height ($H$):

• Definition: The highest point reached by the projectile during its motion.
• Formula:
  $$H=\frac{(v_0\sin \theta {)}^2}{2g}$$

(e) Horizontal Range ($R$):

• Definition: The total horizontal distance covered by the projectile.
• Formula:
  $$R=\frac{v_0^2\sin (2\theta )}{g}$$
• Maximum Range: Achieved when $\theta ={45}^{\circ }$.

3. Assumptions in Projectile Motion

1. Air resistance is negligible.
2. The acceleration due to gravity ($g$) is constant.
3. The motion occurs near the Earth’s surface.

Quick Revision Points

• Trajectory: Parabolic path.
• Horizontal Velocity: Constant ($v_x=v_0\cos \theta$).
• Vertical Velocity: Changes due to gravity ($v_y=v_0\sin \theta -gt$).
• Time of Flight: $T=\frac{2v_0\sin \theta }{g}$.
• Maximum Height: $H=\frac{(v_0\sin \theta {)}^2}{2g}$.
• Horizontal Range: $R=\frac{v_0^2\sin (2\theta )}{g}$.

Previous Year Questions and Answers

• Initial velocity ($v_0$) = $20\text{m/s}$.
• Angle of projection ($\theta$) = ${30}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Time of flight ($T$) = $\frac{2v_0\sin \theta }{g}=\frac{2\cdot 20\cdot \sin ({30}^{\circ })}{9.8}=\frac{2\cdot 20\cdot 0.5}{9.8}=\frac{20}{9.8}\approx 2.04\text{s}$.

• Initial velocity ($v_0$) = $30\text{m/s}$.
• Angle of projection ($\theta$) = ${60}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Maximum height ($H$) = $\frac{(v_0\sin \theta {)}^2}{2g}=\frac{(30\cdot \sin ({60}^{\circ }){)}^2}{2\cdot 9.8}=\frac{(30\cdot 0.866{)}^2}{19.6}=\frac{675.69}{19.6}\approx 34.47\text{m}$.

• Initial velocity ($v_0$) = $15\text{m/s}$.
• Angle of projection ($\theta$) = ${45}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Horizontal range ($R$) = $\frac{v_0^2\sin (2\theta )}{g}=\frac{{15}^2\cdot \sin ({90}^{\circ })}{9.8}=\frac{225\cdot 1}{9.8}\approx 22.96\text{m}$.

• Height ($h$) = $20\text{m}$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Time of flight ($T$) = $\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 20}{9.8}}=\sqrt{\frac{40}{9.8}}\approx \sqrt{4.08}\approx 2.02\text{s}$.

• Initial velocity ($v_0$) = $50\text{m/s}$.
• Angle of projection ($\theta$) = ${45}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Horizontal range ($R$) = $\frac{v_0^2\sin (2\theta )}{g}=\frac{{50}^2\cdot \sin ({90}^{\circ })}{9.8}=\frac{2500\cdot 1}{9.8}\approx 255.1\text{m}$.

Expected Questions

• Initial velocity ($v_0$) = $25\text{m/s}$.
• Angle of projection ($\theta$) = ${37}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Time of flight ($T$) = $\frac{2v_0\sin \theta }{g}=\frac{2\cdot 25\cdot \sin ({37}^{\circ })}{9.8}=\frac{2\cdot 25\cdot 0.6}{9.8}=\frac{30}{9.8}\approx 3.06\text{s}$.

• Initial velocity ($v_0$) = $40\text{m/s}$.
• Angle of projection ($\theta$) = ${60}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Maximum height ($H$) = $\frac{(v_0\sin \theta {)}^2}{2g}=\frac{(40\cdot \sin ({60}^{\circ }){)}^2}{2\cdot 9.8}=\frac{(40\cdot 0.866{)}^2}{19.6}=\frac{1197.16}{19.6}\approx 61.08\text{m}$.

• Height ($h$) = $45\text{m}$.
• Horizontal velocity ($v_x$) = $15\text{m/s}$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Time of flight ($T$) = $\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 45}{9.8}}=\sqrt{\frac{90}{9.8}}\approx \sqrt{9.18}\approx 3.03\text{s}$.
• Horizontal range ($R$) = $v_x\cdot T=15\cdot 3.03\approx 45.45\text{m}$.

• Initial velocity ($v_0$) = $50\text{m/s}$.
• Angle of projection ($\theta$) = ${30}^{\circ }$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Horizontal range ($R$) = $\frac{v_0^2\sin (2\theta )}{g}=\frac{{50}^2\cdot \sin ({60}^{\circ })}{9.8}=\frac{2500\cdot 0.866}{9.8}\approx \frac{2165}{9.8}\approx 220.92\text{m}$.

• Height ($h$) = $10\text{m}$.
• Acceleration due to gravity ($g$) = $9.8{\text{m/s}}^2$.
• Time of flight ($T$) = $\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 10}{9.8}}=\sqrt{\frac{20}{9.8}}\approx \sqrt{2.04}\approx 1.43\text{s}$.