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A student jumps straight up from a diving board that is 3.1 m above the water and hits the water 2.4 seconds later. With what velocity did the student Jump?

Respuesta :

MathPhys

Answer:

10. m/s

Explanation:

Given:

y₀ = 3.1 m

y = 0 m

a = -9.8 m/s²

t = 2.4 s

Find: v₀

y = y₀ + v₀ t + ½ at²

0 m = 3.1 m + v₀ (2.4 s) + ½ (-9.8 m/s²) (2.4 s)²

v₀ = 10.47 m/s

Rounded to two significant figures, the student's initial velocity was 10. m/s.

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