Important Derivations for Class 11 Physics Chapter 4 Motion in a Plane

Important Derivations for Class 11 Physics Chapter 4 Motion in a Plane

Here we are providing important derivations for class 11 Physics Chapter 4 Motion in a Plane. This chapter is also called as Motion in 2D. So here we are presenting derivations for Motion in 2D.

Derivation 1:

Time of Flight, Horizontal Range, Maximum Height and Equation of Path of Projectile in Projectile Motion

Consider a projectile launched with an initial velocity \(v_0\) at an angle \(\theta\) with respect to the horizontal axis. We’ll assume no air resistance and a constant acceleration due to gravity \(g\). The horizontal and vertical components of velocity (\(v_x\) and \(v_y\)) can be calculated as follows: \[ v_x = v_0 \cos(\theta) \quad \text{and} \quad v_y = v_0 \sin(\theta) \] The time of flight (\(T\)) is the total time the projectile stays in the air until it returns to the same vertical position from which it was launched. The time of flight can be found using the vertical motion equation: \[ 0 = v_y \cdot T – \frac{1}{2} g \cdot T^2 \] Solving for \(T\): \[ T = \frac{2 \cdot v_y}{g} = \frac{2 \cdot v_0 \sin(\theta)}{g} \] Now, the horizontal distance (\(R\)) covered by the projectile can be calculated using the time of flight: \[ R = v_x \cdot T = v_0 \cos(\theta) \cdot \frac{2 \cdot v_0 \sin(\theta)}{g} = \frac{2 \cdot v_0^2 \sin(\theta) \cos(\theta)}{g} \] Using the identity \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\), we can simplify: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] Now, let’s find the maximum height (\(H\)) reached by the projectile. The maximum height occurs when the vertical component of velocity becomes zero at the top of the trajectory. We can use the vertical motion equation: \[ v_y = v_0 \sin(\theta) – g \cdot t_{\text{top}} = 0 \] Solving for \(t_{\text{top}}\): \[ t_{\text{top}} = \frac{v_0 \sin(\theta)}{g} \] Now, we can find the maximum height (\(H\)) using the vertical motion equation: \[ H = y_{\text{max}} = y_0 + v_y \cdot t_{\text{top}} – \frac{1}{2} g \cdot t_{\text{top}}^2 \] \[ H = y_0 + v_0 \sin(\theta) \cdot \frac{v_0 \sin(\theta)}{g} – \frac{1}{2} g \cdot \left(\frac{v_0 \sin(\theta)}{g}\right)^2 \] \[ H = y_0 + \frac{v_0^2 \sin^2(\theta)}{g} – \frac{v_0^2 \sin^2(\theta)}{2g} \] \[ H = y_0 + \frac{v_0^2 \sin^2(\theta)}{2g} \] Finally, the equation of the path of the projectile (\(y\) as a function of \(x\)) is a parabolic curve: \[ y = x \tan(\theta) – \frac{g x^2}{2 v_0^2 \cos^2(\theta)} + y_0 \] This equation represents the trajectory of the projectile launched at an angle \(\theta\) from the point \((x_0, y_0)\), neglecting air resistance.

Important Formulas Related to Projectile Motion

1. Horizontal Velocity (\(v_x\)) of the Projectile: \[ v_x = v_0 \cos(\theta) \] 2. Vertical Velocity (\(v_y\)) of the Projectile: \[ v_y = v_0 \sin(\theta) \] 3. Time of Flight (\(T\)): \[ T = \frac{2v_0 \sin(\theta)}{g} \] 4. Maximum Height (\(H\)): \[ H = \frac{v_0^2 \sin^2(\theta)}{2g} \] 5. Range (\(R\)): \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] 6. Vertical Displacement (\(y\)) at Time \(t\): \[ y = v_0 \sin(\theta) t – \frac{1}{2} g t^2 \] 7. Horizontal Displacement (\(x\)) at Time \(t\): \[ x = v_0 \cos(\theta) t \] 8. Equation of the Projectile’s Path: \[ y = x \tan(\theta) – \frac{gx^2}{2v_0^2\cos^2(\theta)} \] where: \(v_0\) = Initial velocity of the projectile
\(\theta\) = Launch angle
\(g\) = Acceleration due to gravity (approximately 9.81 m/s² on Earth)

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