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An astronaut stands on the edge of a lunar crater and throws a half-eaten sandwich horizontally with a velocity of 5.00 m/s. The floor of the crater is 100.0 m below the astronaut. What horizontal distance will the sandwich travel before hitting the floor of the crater? (The acceleration of gravity on the moon is 1.6 m/s².)

a) 312.5 meters
b) 125 meters
c) 50 meters
d) 7.8 meters

User Yanex
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1 Answer

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The half-eaten sandwich thrown horizontally with a velocity of 5.00 m/s will cover a horizontal distance of 62.5 meters on the Moon, before it hits the floor of the crater 100.0 meters below.

The horizontal distance that a half-eaten sandwich will travel before hitting the floor of a lunar crater can be found by determining the time it takes for the sandwich to fall to the craters floor and then using that time to calculate the horizontal distance travelled. To find the time, we can use the second equation of motion s = ut + (1/2)at2 where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Since the sandwich is thrown horizontally, the initial vertical velocity is 0 m/s. Thus, t can be found using the equation 100.0 m = (1/2)(1.6 m/s2)t2. Solving the equation gives us a time t of 12.5 seconds. Next, to find the horizontal distance travelled, we use distance = velocity × time. The initial horizontal velocity is given as 5.00 m/s, so the horizontal distance is 5.00 m/s × 12.5 s, which calculates to 62.5 meters.

So, the sandwich will travel a horizontal distance of 62.5 meters before hitting the floor of the crater.

User Anthony Grist
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