Question

Vertical displacement of bomb on reaching ground​

Answer 1

Since the bomber drops the bomb from a height of 78.4m, the time taken for the bomb to reach the ground will be given by s=ut+21at2 where,

u is the initial velocity in the vertical direction which is 0

s is the vertical displacement from the ground which is −78.4m

a is the acceleration due to gravity g=−9.8ms−2

Upon substituting these values we get t=4s

Since the flight is moving in the horizontal direction with a speed of 150m/s, the distance covered by the bomb in the horizontal direction by the time it reaches the ground is given by (150∗4)m which is equal to 600m

∴The bomb should be released 600m away from the target

Answer 2

Answer:

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Explanation:

Since the bomber drops the bomb from a height of 78.4m, the time taken for the bomb to reach the ground will be given by s=ut+

1/2 at² where,

u is the initial velocity in the vertical direction which is 0

s is the vertical displacement from the ground which is −78.4m

a is the acceleration due to gravity g=−9.8ms

−

2

Upon substituting these values we get t=4s

Since the flight is moving in the horizontal direction with a speed of 150m/s, the distance covered by the bomb in the horizontal direction by the time it reaches the ground is given by (150∗4)m which is equal to 600m

∴The bomb should be released 600m away from the target