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An airline passenger drops a coin while the plane is moving horizontally at 265 m/s.

Randomized Variablesv = 265 m/s
h = 1.9 m
How far does the coin travel horizontally in meters with respect to the Earth if it falls 1.9 m?

User Kartikmaji
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The distance the coin travels horizontally in meters with respect to the Earth, obtained from the time of flight of the coin is about 164.93 meters

The steps used to find the horizontal distance are;

The horizontal velocity of the coin dropped by the airline passenger = 265 m/s

The height the coin falls for which the horizontal distance is required = 1.9 m

The duration the coin falls, t can be found using the following kinetic equation of motion;

s = u·t + (1/2)·g·t²

Where;

s = The vertical distance of travel of the coin

u = The initial vertical velocity, which is zero meters ;per second

t = The duration of flight of the coin

g = The acceleration due to gravity which is about 9.81 m/s²

Therefore;

1.9 ≈ 0 × t + (1/2) × 9.81 × t²

(1/2) × 9.81 × t² = 1.9

t² = 1.9/((1/2) × 9.81)

t ≈ √(1.9/((1/2) × 9.81))

√(1.9/((1/2) × 9.81)) ≈ 0.62 s

The duration of flight, which is the duration the coin falls before reaching a vertical distance of 1.9 meters downwards about 0.62 seconds

Horizontal distance = Horizontal velocity × Duration of travel

The horizontal distance the coin travels is therefore;

265 × √(1.9/((1/2) × 9.81)) ≈ 164.9

The horizontal distance the coin travels is about 164.9 meters

User Randal
by
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