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rocky the flying squirrel is carrying a nut of mass 0.5 kg while flying horizontally at a height of 15 m above the ground at a speed of 12 m/s. bullwinkle is eagerly awaiting the delivery of the nut on the ground. rocky releases the nut as he is directly above bullwinkle. how far from bullwinkle will the nut land if bullwinkle does not move



Answer :

The nut distance from the Bullwinkle after uniform motion is 21 m.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, the parameters given are

m = 0.5 kg

s = 15 m

vx = 12 m/s

vo = 0 m/s

a = g = 9.8 m/s²

Find the time taken of the nut for landing

s = vo . t + 1/2 . a . t²

15 = 0 + 1/2 . 9.8 . t²

t² = 3.06

t = 1.75 s

Find the distance of nut in horizontal direction

vx = x / t

12 = x / 1.75

x = 12 . 1.75

x = 21 m

Find more on uniform motion at: https://brainly.com/question/28040370

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