Ncert Exercises & Solutions

4.24. A fighter plane is flying horizontally at an altitude of 1.5 km with a speed of 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb in order to attack the target?

Hint: Time is taken to travel horizontal and vertical distances will be the same.

Step 1: Draw the diagram.

Consider the adjacent diagram. Let a fighter plane, when it be
at position P, drops a bomb to hit a target T.

Step 2: Calculate the time taken to travel vertical distance.

Altitude of the plane (P'T) = 15 km = 1500 m

$\begin{array}{l} P' T = \frac{1}{2} {gt}^{2} \Rightarrow 1500 = \frac{1}{2} \times 9 .8 t^{2} \\ \Rightarrow t^{2} = \frac{1500}{4 .9} \Rightarrow t = \sqrt{\frac{1500}{49}} = 17 .49 s \end{array}$ $\begin{array}{l} P' T = \frac{1}{2} {gt}^{2} \Rightarrow 1500 = \frac{1}{2} \times 9 .8 t^{2} \\ \Rightarrow t^{2} = \frac{1500}{4 .9} \Rightarrow t = \sqrt{\frac{1500}{49}} = 17 .49 s \end{array}$

Step 3: Write the equation in the horizontal direction and calculate PP'
Speed of the plane $= 720 km / h = 720 \times \frac{5}{18} m / s = 200 m / s$ $= 720 km / h = 720 \times \frac{5}{18} m / s = 200 m / s$

If the bomb hits the target after time t, then the horizontal distance traveled by the bomb.
PP' = $v_{x}$ $v_{x}$ t = 200t

$\begin{array}{l} PP' = 200 \times 17 .49 m \\ Now, tanθ = \frac{P' T}{P' P} = \frac{1500}{200 \times 17 .49} = . 49287 = \tan 23^{\circ} 12' \\ θ = 23^{\circ} 12' . \end{array}$ $\begin{array}{l} PP' = 200 \times 17 .49 m \\ Now, tanθ = \frac{P' T}{P' P} = \frac{1500}{200 \times 17 .49} = . 49287 = \tan 23^{\circ} 12' \\ θ = 23^{\circ} 12' . \end{array}$

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