How High Will a Projectile Rise if Launched at 10.1 km/s?

[bearhug]

At the Earth's surface a projectile is launched straight up at a speed of 10.1 km/s. To what height will it rise?
Universal gravitational constant = 6.673e-11 N m^2/kg^2
Radius of the Earth = 6.370e+6 m
Mass of the Earth = 5.980e+24 kg

I know to use the equation U= - (Gm1m2)/ r
I also need to take into consideration that r is the radius of the Earth and the object is launched at the surface of the earth. However how am I suppose to solve for height when their is no mass of the projectile and what do I do with the speed of the projectile. I originally thought I should consider that U=mgy but that got me nowhere. Any help will be appreciated I'm literally teaching this to myself.

 

[6Stang7]

what do you use potential energy for in physics? what are the units of P.E. what other formual has the same units?

 

[kyle8921]

I can provide an explanation for the height the projectile will reach using the principles of universal gravitation. The equation you mentioned, U = -(Gm1m2)/r, is the potential energy equation for two objects with masses m1 and m2, separated by a distance r, where G is the universal gravitational constant. In this case, m1 represents the mass of the Earth and m2 represents the mass of the projectile.

To find the height the projectile will reach, we can use the conservation of energy principle, which states that the total energy of a system remains constant. In this case, the initial energy of the projectile is all in the form of kinetic energy, which is given by the equation KE = 1/2 mv^2, where m is the mass of the projectile and v is its initial velocity.

At the highest point of the projectile's trajectory, all of its initial kinetic energy will have been converted into potential energy, given by the equation PE = mgh, where m is the mass of the projectile, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and h is the height reached.

Using the conservation of energy principle, we can equate the initial kinetic energy to the final potential energy and solve for h:

1/2 mv^2 = mgh

h = (1/2 v^2)/g

Substituting in the values given:

h = (1/2 * 10.1 km/s)^2 / 9.8 m/s^2

h = 51.25 km

Therefore, the projectile will reach a height of approximately 51.25 km. It is important to note that this calculation assumes that the projectile experiences no air resistance and that the Earth is a perfect sphere. In reality, air resistance and the Earth's shape will affect the height reached by the projectile. I hope this explanation helps and if you have any further questions, I would be happy to assist.