Category: HC Verma Part 1: Mechanics
-
HC Verma Chapter 3 Problem 20 — River crossing: minimum time
Problem Statement A swimmer can swim at 3 m/s in still water. A river 60 m wide flows at 4 m/s. If she aims perpendicular to the bank, find the time to cross and the drift. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles…
-
HC Verma Chapter 3 Problem 19 — Velocity components of horizontal projectile at t
Problem Statement Solve the kinematics problem: A ball is thrown horizontally at 10 m/s. Find its speed after 3 s. ($g = 10$ m/s²) $v_x = 10$ m/s (constant); $v_y = gt$ (downward) $v = \sqrt{v_x^2 + v_y^2}$ Step 1: $v_y = 10 \times 3 = 30$ m/s. Step 2: $v = \sqrt{100 + 900}…
-
HC Verma Chapter 3 Problem 18 — Horizontal projectile from height
Problem Statement Solve the kinematics problem: A ball is thrown horizontally at 15 m/s from a height of 20 m. Find the horizontal distance from the base of the cliff where it lands. ($g = 10$ m/s²) Time to fall: $h = \frac{1}{2}gt^2$; horizontal distance = $v_x \cdot t$ Step 1: $t = \sqrt{2h/g} =…
-
HC Verma Chapter 3 Problem 16 — Time of flight for projectile
Problem Statement Solve the kinematics problem: A ball is projected at 20 m/s at 30°. Find the time of flight. ($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…
-
HC Verma Chapter 3 Problem 17 — Angle for maximum range
Problem Statement At what angle should a projectile be thrown to achieve maximum horizontal range on level ground? Given Information See problem statement for all given quantities. 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…
-
HC Verma Chapter 3 Problem 15 — Projectile: maximum height
Problem Statement Solve the kinematics problem: A ball is projected with speed 20 m/s at 30°. Find the maximum height. ($g = 10$ m/s²) $H = u^2\sin^2\theta/(2g)$ Step 1: $H = \dfrac{400 \times \sin^2 30°}{2 \times 10} = \dfrac{400 \times 0.25}{20} = \dfrac{100}{20} = 5$ m. $$\boxed{H = 5\text{ m}}$$ Given Information $\boxed{H = 5\text{…
-
HC Verma Chapter 3 Problem 14 — Projectile: horizontal range
Problem Statement Solve the kinematics problem: A ball is projected with speed 20 m/s at 30° above horizontal. Find the horizontal range. ($g = 10$ m/s²) $R = u^2\sin 2\theta / g$ Step 1: $R = \dfrac{(20)^2 \sin 60°}{10} = \dfrac{400 \times (\sqrt{3}/2)}{10} = 40\sqrt{3}/2 = 20\sqrt{3} \approx 34.6$ m. $$\boxed{R = 20\sqrt{3} \approx 34.6\text{…
-
HC Verma Chapter 3 Problem 13 — Equation of motion: find s in nth second
Problem Statement A particle starts from rest with acceleration 2 m/s². Find the distance it travels in the 5th second. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion,…
-
HC Verma Chapter 3 Problem 12 — Average speed for two-stage journey
Problem Statement Solve the kinematics problem: A car covers first half of the distance at 40 km/h and second half at 60 km/h. Find the average speed. Average speed = Total distance / Total time Step 1: Let total distance = $2d$. Time for first half: $t_1 = d/40$; for second half: $t_2 = d/60$.…
-
HC Verma Chapter 3 Problem 11 — Relative velocity of two trains (opposite directions)
Problem Statement Solve the kinematics problem: Train A moves at 60 km/h east and train B moves at 40 km/h west. Find the velocity of A relative to B. Opposite directions: $v_{AB} = v_A – v_B = v_A – (-|v_B|) = v_A + |v_B|$ Step 1: Taking east as positive: $v_A = +60$, $v_B =…