HC Verma Chapter 3 Problem 16 — Time of flight for projectile

Problem Statement

Solve the kinematics problem: A ball is projected at 20 m/s at 30°. Find the time of flight. ($g = 10$ m/s²) $T = 2u\sin\theta / g$ Step 1: $T = \dfrac{2 \times 20 \times \sin 30°}{10} = \dfrac{2 \times 20 \times 0.5}{10} = \dfrac{20}{10} = 2$ s. $$\boxed{T = 2\text{ s}}$$

Given Information

  • $T = 2u$
  • $\boxed{T = 2\text{ s}$

Physical Concepts & Formulas

Projectile motion decomposes into independent horizontal and vertical components. Horizontal: constant velocity (no air resistance). Vertical: constant downward acceleration $g$. The trajectory is a parabola. Maximum range occurs at $45°$ launch angle; max height at $90°$.

  • $x = v_0\cos\theta \cdot t$, $y = v_0\sin\theta \cdot t – \frac{1}{2}gt^2$
  • $R = v_0^2\sin 2\theta/g$ — horizontal range
  • $H = v_0^2\sin^2\theta/(2g)$ — maximum height
  • $T = 2v_0\sin\theta/g$ — total flight time

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\boxed{T = 2\text{ s}}$$

$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$

$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$

Answer

$$\boxed{T = 2\text{ s}}$$

Physical Interpretation

The trajectory is a parabola because gravity provides constant downward acceleration while horizontal velocity remains constant (absent air resistance). Real projectiles deviate due to drag, especially at high speeds.


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