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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 $g = 10 m s^{-2}$ ).
Solution :
Step 1: Coordinate System and Unit Conversion
Convert all given variables into SI units for consistency.
- Altitude of the plane ($h$): $1.5\text{ km} = 1500\text{ m}$
- Speed of the plane ($v_p$): $720\text{ km/h} = 720 \times \left(\frac{5}{18}\right) = 200\text{ m/s}$
- Muzzle speed of shell ($u$): $600\text{ m/s}$
- Acceleration due to gravity ($g$): $10\text{ m/s}^2$
Step 2: Kinematic Equations for the Shell and Plane
Let the gun be at the origin $(0, 0)$. The plane is at $(0, h)$ at $t=0$.
The position of the shell at time $t$ is:
$x_s = (u \sin \theta)t$
$y_s = (u \cos \theta)t - \frac{1}{2}gt^2$
The position of the plane at time $t$ is:
$x_p = v_p t$
$y_p = h$
[Per the Principle of Independent Horizontal and Vertical Motion]
Step 3: Solving for the Angle $\theta$
To hit the plane, $x_s = x_p$ and $y_s = y_p$ at the same time $t$.
From $x_s = x_p$: $(u \sin \theta)t = v_p t \implies \sin \theta = \frac{v_p}{u}$
$\sin \theta = \frac{200}{600} = \frac{1}{3} \approx 0.3333$
$\theta = \arcsin(0.3333) \approx 19.47^\circ$
[By Trigonometric Ratio Definition]
Step 4: Verification of Time and Vertical Position
Calculate $t$ from $y_s = y_p$:
$u \cos \theta \cdot t - \frac{1}{2}gt^2 = h$
$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (1/3)^2} = \sqrt{8/9} = \frac{2\sqrt{2}}{3} \approx 0.9428$
Substituting values: $600 \times \left(\frac{2\sqrt{2}}{3}\right) t - 5t^2 = 1500$
$400\sqrt{2} t - 5t^2 = 1500 \implies 5t^2 - 565.68t + 1500 = 0$
Using the quadratic formula, $t \approx 2.7\text{ s}$ (or $110\text{ s}$, which is impractical). Since the shell reaches the altitude $h$ within physical limits, the angle is valid.
Step 5: Calculating Minimum Altitude ($h_{min}$)
The shell reaches its maximum vertical displacement when its vertical velocity component $v_y = 0$. The maximum vertical height $H_{max}$ is given by: $H_{max} = \frac{(u \cos \theta)^2}{2g}$ $H_{max} = \frac{(600 \times \frac{2\sqrt{2}}{3})^2}{2 \times 10} = \frac{(400\sqrt{2})^2}{20} = \frac{160000 \times 2}{20} = 16000\text{ m}$ [Per the Law of Conservation of Energy and Kinematic Constraints]
Final Solution: The angle from the vertical is $19.47^\circ$. The minimum altitude to avoid being hit is $16000\text{ m}$ ($16\text{ km}$).
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.
- Q4.31: 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 $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{ox}}\right)$
- 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.
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