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Lemur [1.5K]
3 years ago
9

A toy projectile is fired from the ground vertically upward with an initial velocity of 26.5 m/s. The project arrives at its max

imum altitude in 2.7s.
What is the velocity of the projectile when it hits the ground? How?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0
<span>26.5 m/s Ignoring air resistance, the projectile will have an initial kinetic energy expressed by E = 0.5 M V^2 and as the projectile travels upward, it's kinetic energy will decrease while it's potential gravitational kinetic energy increases such that the sum of both is constant. Then once it starts to descent, the potential energy is converted back into kinetic energy. And when it finally reaches the ground, it's kinetic energy will exactly match the original kinetic energy it had initially. And the only way it can match since the mass never changes is for it to have the same velocity as it had when it was first fired.</span>
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Your friend thinks that the escape speed should be greater for more massive objects than for less massive objects. Provide an ar
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Concepts and Principles

1- Kinetic Energy: The kinetic energy of an object is:

K=1/2*m*v^2                                                         (1)  

where m is the object's mass and v is its speed relative to the chosen coordinate system.  

2- Gravitational potential energy of a system consisting of Earth and any object is:  

 U_g = -Gm_E*m_o/r*E-o                                   (2)  

where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.  

Solution  

The argument:  

My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.  

The counterargument:  

We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):

K=1/2*m*v^2                                                         (1)  

and a gravitational potential energy Ug given by Equation (2):  

Ug = -G*Mm/R

where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:  

K+U=0  

1/2*m*v^2 + (-G*Mm/R) = 0

1/2*m*v^2 =  G*Mm/R

1/2*v^2 = G*M/R

solving for v we get

v = √2G*M/R

so we see v does not depend on the mass of the projectile

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