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Q4.32(b):
(b) Shows that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$ where the symbols have their usual meaning.
Solution :
Step 1: Establishing Fundamental Kinematic Equations
For a projectile launched from the origin $(0,0)$ with an initial velocity $v_0$ at an angle $\theta_0$ to the horizontal, the equations of motion are defined as follows:
- The maximum height ($h_m$) reached by the projectile is given by: $h_m = \frac{v_0^2 \sin^2 \theta_0}{2g}$ [Per the kinematic equation $v_y^2 = u_y^2 - 2gh$ at peak where $v_y = 0$]
- The horizontal range ($R$) of the projectile is given by: $R = \frac{v_0^2 \sin(2\theta_0)}{g} = \frac{2v_0^2 \sin \theta_0 \cos \theta_0}{g}$ [By the trigonometric identity $\sin(2\theta_0) = 2 \sin \theta_0 \cos \theta_0$]
Step 2: Constructing the Ratio of $h_m$ to $R$
To isolate $\theta_0$, we compute the ratio $\frac{h_m}{R}$:
$\frac{h_m}{R} = \frac{\left( \frac{v_0^2 \sin^2 \theta_0}{2g} \right)}{\left( \frac{2 v_0^2 \sin \theta_0 \cos \theta_0}{g} \right)}$
Step 3: Algebraic Simplification
Cancel the common terms $v_0^2$ and $g$ from the numerator and denominator:
$\frac{h_m}{R} = \frac{\sin^2 \theta_0}{2} \cdot \frac{1}{2 \sin \theta_0 \cos \theta_0}$
$\frac{h_m}{R} = \frac{\sin^2 \theta_0}{4 \sin \theta_0 \cos \theta_0}$
Canceling one $\sin \theta_0$ term:
$\frac{h_m}{R} = \frac{\sin \theta_0}{4 \cos \theta_0}$
Using the identity $\tan \theta_0 = \frac{\sin \theta_0}{\cos \theta_0}$:
$\frac{h_m}{R} = \frac{1}{4} \tan \theta_0$
Step 4: Isolating the Projection Angle $\theta_0$
Multiply both sides by 4:
$\tan \theta_0 = \frac{4h_m}{R}$
Applying the inverse tangent function to both sides [By definition of inverse trigonometric functions]:
$\theta_0 = \tan^{-1} \left( \frac{4h_m}{R} \right)$
Visual Representation of Projectile Trajectory
Final Solution: $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$
More Questions from Class 11 Physics Motion in a Plane Additional Exercises
- Q4.26: A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer.
- Q4.27: A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ?
- Q4.28(a): Can you associate vectors with (a) the length of a wire bent into a loop,
- Q4.28(b): Can you associate vectors with (b) a plane area,
- Q4.28(c): Can you associate vectors with (c) a sphere ? Explain.
- Q4.29: A bullet fired at an angle of $30^{\circ}$ with the horizontal hits the ground $3.0 km$ away. By adjusting its angle of projection, can one hope to hit a target $5.0 km$ away ? Assume the muzzle speed to be fixed, and neglect air resistance.
- Q4.30: A fighter plane flying horizontally at an altitude of $1.5 km$ with speed $720 km/h$ passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed $600 m s^{-1}$ to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take $g = 10 m s^{-2}$ ).
- Q4.31: A cyclist is riding with a speed of $27 km/h$. As he approaches a circular turn on the road of radius $80 m$, he applies brakes and reduces his speed at the constant rate of $0.50 m/s$ every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?
- Q4.32(a): (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{ox}}\right)$
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