Question

A bomb is dropped on an enemy post by an aeroplane
with a horizontal velocity of 60 km/hr and at a height of 490
m. How far the aeroplane must be from the enemy
me time of dropping the bomb, so that it may directly hit the
target? (g = 9.8 m/s2) ......pls answer asap​

Answer 1

Answer:

answer is 166.67 m

Explanation:

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Answer 2

The aeroplane travel a distance of 166.6 m at time of dropping the bomb.

Explanation:

Given that,

Velocity of the aeroplane, v = 60 km/h = 16.66 m/s

Height, h = 490 m

Let t is the time taken by aeroplane. Using second equation of motion to find it as :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\t=\sqrt{\dfrac{2\times 490}{9.8}} \\t=10\ s$

Let the aeroplane travel a distance of x m at time of dropping the bomb. Distance traveled is given by :

$x=v\times t\\\\x=16.66\ m/s\times 10\ s\\\\x=166.6\ m$

So, the aeroplane travel a distance of 166.6 m at time of dropping the bomb.

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