at the earths surface, a projectile is launched straight up at a speed of 10km/. To what height will it rise?

i need some major help here please don't know where to begin on this one!

Do you mean 10 km/s?

Solve the energy equation
(1/2) m V^2 - G mM/Re = -GmM/(Re + h)

Re is the earth's radius and h is the maximum altitude. V is the initial velocity. The -GmM tems are the poential energy in the inverse-square gravitaional field of the Earth. M is the Earth's mass.

Ig you don't want to have to look up G, remember that
GM/Re^2 = g, the acceleration of gravity at the Earth's surface.

The first equation can be rewritten, after cancelling out m's,
(1/2) V^2 = g Re [1 - 1/(1 + h/Re)]

Solve for h, but check my math first

To determine the maximum height reached by a projectile launched straight up, we can use basic principles of kinematics.

First, it's important to know the initial velocity of the projectile, which is given as 10 km/s in this case. However, it's helpful to convert this value to the appropriate unit of meters per second (m/s) to maintain consistency in the calculations.

To convert km/s to m/s, we need to multiply the given velocity by 1000 due to the conversion factor: 1 km = 1000 m. Therefore, the initial velocity of the projectile is 10 km/s * 1000 m/km = 10,000 m/s.

Now, we can identify the key information necessary to solve this problem:
- The initial velocity (u) = 10,000 m/s
- The acceleration due to gravity (g) = 9.8 m/s^2 (assuming the motion occurs near the Earth's surface)
- The final velocity (v) at the highest point is 0 m/s, as the projectile momentarily stops before falling back down.
- We need to find the maximum height (h).

To begin, we can use the equation of motion relating the final velocity (v), initial velocity (u), acceleration (a), and displacement (s):
v^2 = u^2 + 2as

Since the projectile reaches its highest point, the final velocity (v) is 0. Substituting the known values, the equation becomes:
0^2 = (10,000 m/s)^2 + 2 * a * h

Simplifying, we get:
0 = 100,000,000 m^2/s^2 + 2ah

Rearranging the equation, we find:
2ah = -100,000,000 m^2/s^2
ah = -50,000,000 m^2/s^2

Finally, we need to consider the acceleration due to gravity (g) acting in the opposite direction to the displacement (h). Thus, g is negative in this equation: ah = -50,000,000 m^2/s^2.

Substituting the value for acceleration due to gravity (g = -9.8 m/s^2), we can solve for the maximum height (h):
-9.8 m/s^2 * h = -50,000,000 m^2/s^2
h = (-50,000,000 m^2/s^2) / (-9.8 m/s^2)

Calculating this value, we find:
h ≈ 5,102,040 meters

Therefore, the projectile will rise to approximately 5,102,040 meters or 5,102.04 kilometers above the Earth's surface.