An aeroplane flying horizontally with a speed of 360 km h-1 releases a bomb at a height of 490 m from the ground. If g = 9.8 m s-2, it will strike the ground ata)10 kmb)100 kmc)1 kmd)16 kmCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

To solve this problem, we can use the equations of motion under constant acceleration.

Given:
Initial velocity (u) = 360 km/h
Height (h) = 490 m
Acceleration due to gravity (g) = 9.8 m/s^2

Converting the initial velocity from km/h to m/s:
u = 360 km/h = (360 * 1000) m/3600 s = 100 m/s

We need to find the time taken for the bomb to hit the ground. We can use the equation:
h = ut + (1/2)gt^2

Rearranging the equation to solve for time (t):
(1/2)gt^2 + ut - h = 0

Substituting the given values:
(1/2)(9.8)t^2 + 100t - 490 = 0

This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)

Substituting the values for a, b, and c:
t = (-(100) ± √((100)^2 - 4(1/2)(-9.8)(-490)))/(2(1/2)(9.8))

Simplifying the equation:
t = (-100 ± √(10000 + 9608))/(9.8)

t = (-100 ± √(19608))/(9.8)

Since time cannot be negative, we take the positive value:
t = (-100 + √(19608))/(9.8)

Calculating the value inside the square root:
√(19608) = 140

t = (-100 + 140)/(9.8) = 40/9.8 ≈ 4.08 s

The time taken for the bomb to hit the ground is approximately 4.08 seconds.

To find the distance traveled by the plane during this time, we can use the equation:
s = ut + (1/2)at^2

Substituting the values:
s = (100)(4.08) + (1/2)(9.8)(4.08)^2

Calculating the value:
s = 408 + 80 ≈ 488 m

The distance traveled by the plane during this time is approximately 488 meters.

Therefore, the bomb will strike the ground at a distance of approximately 488 meters from the release point. This corresponds to option 'c', 1 km.

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