Category: HC Verma Part 1: Mechanics
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HC Verma Chapter 3 Problem 50 — Projectile speed at a given height
Problem Statement Solve the kinematics problem: A projectile is fired at $u = 20$ m/s, $\theta = 60°$. Find the speed when the vertical component of velocity equals the horizontal component. ($g = 10$ m/s²) $v_x = u\cos\theta$; $v_y = u\sin\theta – gt$; condition $v_y = v_x$ Step 1: $v_x = 20\cos 60° = 10$…
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HC Verma Chapter 3 Problem 49 — Resultant velocity of boat in current
Problem Statement Analyze the circuit: A boat can travel at 8 m/s in still water. It crosses a river 160 m wide with a current of 6 m/s. If the boat heads directly across, find the actual speed and direction of the boat and the time to cross. Resultant velocity = vector sum of boat…
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HC Verma Chapter 3 Problem 48 — Uniform circular motion: period
Problem Statement A particle completes one revolution in a circle of radius 2 m in 4 s. Find the linear speed and centripetal acceleration. Given Information See problem statement for all given quantities. Physical Concepts & Formulas Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r =…
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HC Verma Chapter 3 Problem 46 — Magnitude of displacement vector in 2D
Problem Statement Solve the kinematics problem: A particle moves from point $(1, 2)$ to $(4, 6)$ m. Find the magnitude and direction of the displacement. $\vec{\Delta r} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j}$ Step 1: $\vec{\Delta r} = (4-1)\hat{i}+(6-2)\hat{j} = 3\hat{i}+4\hat{j}$ m. Step 2: $|\vec{\Delta r}| = \sqrt{9+16} = 5$ m. Step 3: Angle: Given Information See…
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HC Verma Chapter 3 Problem 47 — Acceleration of particle in x-y plane
Problem Statement Solve the kinematics problem: A particle moves with $v_x = 6t$ and $v_y = 4$ m/s (constants given at $t$). Find the acceleration at $t = 2$ s. $a_x = dv_x/dt$, $a_y = dv_y/dt$ Step 1: $a_x = 6$ m/s² (constant); $a_y = 0$. Step 2: $\vec{a} = 6\hat{i}$ m/s² (independent of $t$).…
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HC Verma Chapter 3 Problem 45 — Equations of motion for constant acceleration
Problem Statement Solve the kinematics problem: Derive the three kinematic equations for constant acceleration starting from the definition $a = dv/dt$. Integration; $a = $ constant Eq. 1: $dv/dt = a \Rightarrow v = u + at$. Eq. 2: $s = \displaystyle\int_0^t v\,dt = \int_0^t (u+at)\,dt = ut + \tfrac{1}{2}at^2$. Eq. 3: Eliminate $t$: from…
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HC Verma Chapter 3 Problem 44 — Average acceleration in projectile
Problem Statement Solve the kinematics problem: A projectile is fired at 20 m/s at 60° above horizontal. Find the average acceleration over the first 2 s. ($g = 10$ m/s²) $\vec{a}_{avg} = \Delta\vec{v}/\Delta t$; for projectile under gravity, $\vec{a} = -g\hat{j}$ always Step 1: The acceleration is constant: $\vec{a} = -10\hat{j}$ m/s² at all tim…
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HC Verma Chapter 3 Problem 43 — Instantaneous velocity direction
Problem Statement Solve the kinematics problem: A particle has $x = 3t$ and $y = 4t^2$ (SI). Find the direction of velocity at $t = 2$ s. $v_x = dx/dt$, $v_y = dy/dt$; direction $= \arctan(v_y/v_x)$ Step 1: $v_x = 3$ m/s; $v_y = 8t$. At $t=2$: $v_y = 16$ m/s. Step 2: $\tan\alpha =…
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HC Verma Chapter 3 Problem 42 — Range at complementary angles
Problem Statement Show that two projectile angles $\theta$ and $(90° – \theta)$ give the same 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…
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HC Verma Chapter 3 Problem 41 — Body thrown up from moving vehicle
Problem Statement A ball is thrown vertically upward at 10 m/s from a train moving horizontally at 20 m/s. Find the speed and angle of projection as seen from the ground. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The…