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seropon [69]
3 years ago
10

A 60-kilogram man holding a 20-kilogram box rides on a skateboard at a speed of positive 7 meters per second. He throws the box

behind him, giving it a velocity of negative 5 meters per second, with respect to the ground. What is his velocity, in meters per second, after throwing the object?
Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

v_1=11\ m/s

Explanation:

It is given that,

Mass of the man, m₁ = 60 kg

Mass of the box, m₂ = 20 kg

Speed of man+ box, V = +7 m/s

Speed of the box, when it was thrown, v₂ = -5 m/s

To find,

Velocity of the man after throwing the box

Solution,

Let v₁ be the velocity of the man after throwing the box. By using the conservation of momentum, it can be calculated.

(m_1+m_2)V=m_1v_1+m_2v_2

(60+20)\times 7=60\times v_1+20\times (-5)

v_1=11\ m/s

Therefore, the velocity of the man after throwing the object is 11 m/s towards his initial direction.

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Start with what you know. In the y-direction, we know the jumper must fall 15 meters and starts with 0 velocity in the y direction. You can also assume that the acceleration of gravity is pulling down on the jumper at 9.8 m/s. Once you have these three you can plug it into kinematic a equation to find time

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Suppose you ride your bike to the library traveling at 0.5 km/min. It takes you 25 minutes to get to the library. How far did yo
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