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An inflated ball on the surface of Mars bounces off at an angle of 45.0° with a velocity of 15.34 meters/second. What is the resultant velocity with which the ball hits the surface of Mars at the end of its parabolic path?
(gmars = 3.75 meters/second2)

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NobleAlexander
Oh I'm an idiot. ;-; 
The answer is B. Let me explain why! 

For any given parabolic path the velocity and angle will be the same at the start and the end. 
The only variable which will be different from the start is the distance/position. 

Imagine it like this you have a ball and you throw it upward. The only force that pulls it back down is gravity. (Well.. if you factor out air resistance, although for a ball that'd be quite negligible) When it hits the ground it begins to bounce again. The reason the bounce declines isn't because the velocity declined due to gravity. It's a result of energy being transferred into the ground. 

I hope that helped somewhat. :p 

Answer:

Velocity, v = 15.34 "m

Explanation:

It is given that, An inflated ball on the surface of Mars bounces off at an angle of 45.0° with a velocity of 15.34 meters/second.

The inflated ball acts like as a projectile. The motion shown by the projectile is called its projectile motion.

The velocity and the angle of projection in case of projectile remains the same. Hence, this is the required solution.

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