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nignag [31]
2 years ago
11

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

Physics
1 answer:
Darina [25.2K]2 years ago
7 0

Answer:

<em>a. t = 2.02 s </em>

<em>b. d = 20.2 m</em>

Explanation:

<u>Horizontal Motion </u>

If an object is thrown horizontally from a height h with a speed v, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The time the object takes to hit the ground can be calculated as follows:

\displaystyle t=\sqrt{\frac{2h}{g}}

The time does not depend on the initial speed.

The range or maximum horizontal distance traveled by the object can be calculated by the equation:

\displaystyle d=v.t

The man standing on the edge of the h=20 m cliff throws a rock with an initial horizontal speed of v=10 m/s.

a.

The time taken by the rock to reach the ground is:

\displaystyle t=\sqrt{\frac{2*20}{9.8}}

\displaystyle t=\sqrt{4.0816}

t = 2.02 s

b.

The range is:

\displaystyle d=10\cdot 2.02

d = 20.2 m

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