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A rock is thrown straight downward from a tree limb with an initial velocity v0. The rock has constant downward acceleration of 10 m/s2 during its fall. The rock moves 7 times as far during the first 3 seconds as during the first 1 second of fall. What is the value (magnitude only) of v0?

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Answer:

v₀ = 15 m/s

Step-by-step explanation:

given,

initial velocity = v₀

down acceleration of rock = 10 m/s²

rock distance

S₄ = 7 x S₁

From kinematic equations

S = v₀ t+0.5 at²

at t = 1 s

S₁ = v₀ (1)+0.5 x 10 x 1²

S₁ = v₀+ 5......(1)

at t = 4 s

S₄ = v₀ (4)+0.5 x 10 x 4²

S₄ = 4 v₀+80.....(2)

from equation (1) and (2)

7( v₀ + 5 ) = 4 v₀ +80

3 v₀ = 80 - 35

3 v₀ = 45

v₀ = 15 m/s

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