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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 after 2.8 seconds? what is the bullet's velocity at this time? (accelaration of 9.8m/s)

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MathPhys

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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