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DanielleElmas [232]
3 years ago
15

if a bullet fired beneath the water, straight up, breaks through the surface at a speed of 25m/s, how high will the bullet be af

ter 2.8 seconds? what is the bullet's velocity at this time? (accelaration of 9.8m/s)
Physics
1 answer:
Doss [256]3 years ago
8 0

Answer:

32 m and -2.4 m/s

Explanation:

Given:

v₀ = 25 m/s

t = 2.8 s

a = -9.8 m/s²

Find: Δy, v

Δy = v₀ t + ½ at²

Δy = (25 m/s) (2.8 s) + ½ (-9.8 m/s²) (2.8 s)²

Δy = 31.6 m

v = at + v₀

v = (-9.8 m/s²) (2.8 s) + 25 m/s

v = -2.44 m/s

Rounded to two significant figures, the bullet reaches a height of 32 m and a velocity of -2.4 m/s.

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<em>v = 381 m/s</em>

Explanation:

<u>Linear Speed</u>

The linear speed of the bullet is calculated by the formula:

\displaystyle v=\frac{x}{t}

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x = Distance traveled

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The formula for the angular speed of a rotating object is:

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\displaystyle t=\frac{\theta}{\omega}

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