You are exploring a distant planet. When your spaceship is in a circular orbit a

Question

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surface, the ship's orbital speed is 5500 m/s . By observing the planet, you determine its radius to be 4.48×10^6 m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 13.6 m/s at an angle of 30.8 above the horizontal. If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile? Please explain process

Explanation / Answer

Range of a Projectile is givne by

R = uo^2 sin 2 theta /g

here g is the accleration due to gravity

g is obtained from V^2 = 2gs

g = (5500*5500)/(2* (4.48e 6 + 630000))

g = 2.95 m/s^2

so

Range = 13.6*13.6 * sin (2 * 30.8)/2.95

R = 55.15 m

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