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guajiro [1.7K]
3 years ago
10

An object is thrown upwards with a velocity of 3 meters per second, such that the only force acting on it is gravity (accelerati

on is -9.8). What will the object's velocity be after 2 seconds?
A.) -16.6 meters per second (down)
B.) 16.6 meters per second (up)
C.) -22.6 meters per second (down)
D.) 22.6 meters per second (up)
E.) 3 meters per second (up)
Physics
2 answers:
3241004551 [841]3 years ago
3 0

Answer:

Correct answer:  A.)  V = - 16.6 m/s down

Explanation:

Given:

V₀ = 3 m/s  initial velocity

t = 2 seconds

g = 9.8 m/s²

V(t) = V(2) = ?

The movement described is a vertical upward shot

For velocity at any time is valid the next formula

V = V₀ - g · t

V = 3 - (9.8 · 2) = 3 - 19.6 = - 16.6 m/s down

Under condition that it has a enough drop height with respect to the ejection point.

God is with you!!!

NNADVOKAT [17]3 years ago
3 0
The answer is A. DOWN
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Question:
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Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

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m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

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ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE      (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

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ME = KE + PE      (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

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