Question

A bomb is dropped from an aeroplane when it is directly above the target at a height

of 2000 m. The aeroplane is moving with a horizontal speed of 540 kmh-1. By how much distance will the bomb miss the target?​

Answer 1

Answer:-

• The bomb will miss the target by 3 km.

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• Given:-

Height from which the bomb is dropped = 2000 m

Speed of the aeroplane = 540 kmh-¹

→ $540 \times \dfrac{5}{18}$

→ 150 m/s

• To Find:-

Distance by which the bomb will miss the target = ?

• Solution:-

Time taken to reach the ground =$\sf{\sqrt{\dfrac{(2h)}{g}}}$

Taking g = 10 m/s

→ $\sqrt{\dfrac{2 \times 2000}{10}}$

→ $\sqrt{2 \times 200}$

→ $\sqrt{400}$

→ $\bf{20 \: sec}$

• Horizontal distance = Horizontal Speed × Time of fall

→ 150 m/s × 20 sec

→ 3000 m

→ 3 km

Therefore, the bomb will miss the target by 3 km.

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Some Formulas Related to PROJECTILE:-

• Time period:-

T = $\sf{\sqrt{\dfrac{(2h)}{g}}}$

T = $\sf{\dfrac{2u sin \theta}{g}}$ [Angular]

• Horizontal Range:-

R = $\sf{u{\sqrt{\dfrac{(2h)}{g}}}}$

R = $\sf{\dfrac{u^2 sin 2 \theta}{2g}}$ [Angular]

• Maximum Height:-

$H_{max.}$ = $\sf{\dfrac{u^2 sin^2 \theta}{2g}}$

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