Question

An astronaut stands on the edge of a lunar crater and throws a half-eaten Twinkie™ 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 Twinkie™ travel before hitting the floor of the crater? (The acceleration of gravity on the moon is 1/6th that of the Earth).

Answer 1

33.35 meters

Step-by-step explanation:

The horizontal distance traveled by the Twinkie™ can be calculated using the following equation:

d = v0 * t

Where v0 is the initial horizontal velocity, and t is the time it takes for the Twinkie™ to reach the floor of the crater. To find t, we can use the equation for the vertical motion of the Twinkie™:

y = v0y * t - (1/2) * g * t^2

Where v0y is the initial vertical velocity, g is the acceleration due to gravity on the moon (1/6th that of the Earth), and y is the vertical distance from the astronaut to the floor of the crater (100 m). Since the Twinkie™ was thrown horizontally, the initial vertical velocity is 0, so the equation simplifies to:

100 = - (1/2) * (1/6 * 9.8) * t^2

Solving for t:

t = sqrt(100 * 2 / (1/6 * 9.8)) = sqrt(100 * 2 * 6 / 9.8) = sqrt(1200 / 9.8) = 6.67 seconds

Finally, we can use the horizontal velocity and the time to calculate the horizontal distance traveled by the Twinkie™:

d = v0 * t = 5.00 m/s * 6.67 s = 33.35 m

So the Twinkie™ will travel 33.35 meters horizontally before hitting the floor of the crater.