Question

An aeroplane flying horizontally at an altitude of 490 m with a speed of 180 kmph drops a bomb. The
horizontal distance at which it hits the ground is

a) 500 m
b) 1000 m
c) 250 m
d) 50 m​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Height of the plane from ground (H) = 490m.
• Speed of the bomb (u) = 180 km/h.
• Let the horizontal distance be = "R".

$\huge{\bold{\underline{Explanation:-}}}$

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Speed needs to be converted into m/s. (S.I units)

Now,

$\large{\tt u = 180 \: km/h = 180 \times \dfrac{5}{18}}$

$\large{\tt u = 180 \: km/h = \cancel{180} \times \dfrac{5}{\cancel{18}}}$

Now,

$\large{\boxed{\tt u = 50 \: m/s}}$

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As the bomb is projected horizontally the vertical component of velocity will be Zero. and The horizontal component will be equal to the Initial velocity of the bomb.

• I.e. ${\tt u_x = 50 \: m/s}$
• ${\tt u_y = 0 \: m/s}$

Now, Applying second kinematic equation for Vertical motion.

$\large{\boxed{\tt H = u_yt + \dfrac{1}{2}a_yt^2}}$

Substituting the values,

$\large{\tt 490 = 0 + \dfrac{1}{2} \times 9.8 \times t^2}$

$\large{\tt 490 = \dfrac{1}{\cancel{2}} \times \cancel{9.8} \times t^2}$

$\large{\tt 490 = 4.9 \times t^2}$

$\large{\tt t^2 = \dfrac{490}{4.9}}$

It becomes,

$\large{\tt t^2 = \dfrac{\cancel{490}}{\cancel{4.9}}}$

$\large{\tt t^2 = 100}$

$\large{\tt t = \sqrt{100}}$

$\large{\boxed{\tt t = 10 \: Seconds}}$

Therefore, the time taken by the bomb to reach the ground is 10 seconds.

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Now, Applying Second kinematic equation for Horizontal motion.

$\large{\boxed{\tt R = u_xt + \dfrac{1}{2}a_xt^2}}$

Here the Horizontal acceleration will be Zero.(As it is projected horizontally)

Substituting the values,

$\large{\tt R = 50 \times 10 + \dfrac{1}{2} \times 0 \times (10)^2}$

$\large{\tt R = 500 + 0}$

$\huge{\boxed{\boxed{\tt R = 500 \: meters}}}$

So, the Horizontal distance at which the bomb strikes is 500 meters.

Therefore, option- (a) is correct.

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