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A man throws a pebble upward from a point on a cliff 85 m above the sea. The pebble drops into the water 6.2 s after it is thrown. Assuming that air resistance is negligible, what is the velocity of the pebble just after it is thrown?
A) 11 m/s
B) 14 m/s
C) 17 m/s
D) 20 m/s
Pleease help with this i honestly dont understand it..

Respuesta :

MathPhys

Answer:

C) 17 m/s

Explanation:

Given:

y₀ = 85 m

y = 0 m

t = 6.2 s

a = -9.8 m/s²

Find: v₀

y = y₀ + v₀ t + ½ at²

(0 m) = (85 m) + v₀ (6.2 s) + ½ (-9.8 m/s²) (6.2 s)²

v₀ = 16.7 m/s

Rounded to two significant figures, the initial velocity is 17 m/s.

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