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

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- when it be at position P, drops a bomb to hit a target T

Let $\theta$

Speed of the plane =720km/h = 720 $\times 5/18$ = 200m/s

Altitude of the plane P’T = 1.5km= 1500m

If bomb hits the target after time t then horizontal distance travelled by the bomb PP’=u $\times t$ =200t

Vertical distance travelled by the bomb P’T=1/2gt²

1500 = ½ (9.8)t²

So t²= 1500/4.9, t = $\sqrt[]{\frac{1500}{4.9}}=17.49s$

PP’=200 (17.49)m=

tan $\theta$ =P’T/P’P=1500/200 (17.49)=0.49287= tan23⁰12’

 

<p>This is a Short Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation</strong>- when it be at position P, drops a bomb to hit a target T</p><p>Let &lt;p’pt=<span contenteditable="false"> <math> <mi>θ</mi> </math> &lt;/p’pt=</span></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745403311phprCsylP_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745403311phprCsylP.jpeg" alt="" width="161" height="129"></picture></div><div><p>Speed of the plane =720km/h = 720<span contenteditable="false"> <math> <mo>×</mo> <mn>5</mn> <mo>/</mo> <mn>18</mn> </math> </span>= 200m/s</p><p>Altitude of the plane P’T = 1.5km= 1500m</p><p>If bomb hits the target after time t then horizontal distance travelled by the bomb PP’=u<span contenteditable="false"> <math> <mo>×</mo> <mi>t</mi> </math> </span>=200t</p><p>Vertical distance travelled by the bomb P’T=1/2gt²</p><p>1500 = ½ (9.8)t²</p><p>So t²= 1500/4.9, t =<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1500</mn> </mrow> </mrow> <mrow> <mrow> <mn>4.9</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>17.49</mn> <mi>s</mi> </math> </span></p><p>PP’=200 (17.49)m=</p><p>tan<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>=P’T/P’P=1500/200 (17.49)=0.49287= tan23⁰12’</p></div></div><p>&nbsp;</p>

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