At the earthâ€™s surface, a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise?
Ignore air resistance and the rotation of the Earth.

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This is a simple kinematics problem. First we look at the equation for the velocity of the projectile: v = v0 + at. v0 is the initial velocity (10,000 m/s), and a is the acceleration (-9.81 m/s^2, the acceleration due to gravity). Thus we have v(t) = 10000 - 9.81t When the projectile is at its peak its velocity is zero, so we solve the previous equation for v(t) = 0, giving us t = 1019.37 s.

Next, we find the height. The equation for the position of the projectile is x(t) = v0*t + (1/2)*a*t^2. We want to find x(t = 1019.37). The result is 5.097 * 10^6 m, or 5,097 km.

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