An astronaut stands on the edge of a lunar crater and throws a half-eaten Twinkie™ horizontally with a velocity of 2.2 m/s. The floor of the crater is 95.4 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).
i got a time of 4.41 and then multiplied it by the 2.2 velocity and got 9.71
your time is wrong. You used g for Earth, not the moon.
To find the horizontal distance the Twinkie™ will travel before hitting the floor of the crater, we can use the equations of motion.
First, we need to find the time it takes for the Twinkie™ to fall to the floor of the crater. We can use the formula for the time of flight in free fall, which is derived from the kinematic equation for vertical motion:
t = sqrt((2h) / g),
where
t is the time of flight,
h is the height of the crater (95.4 m in this case),
and g is the acceleration due to gravity on the moon (1/6th the acceleration due to gravity on Earth, or approximately 1.6 m/s²).
Plugging in the values:
t = sqrt((2 * 95.4) / 1.6) = sqrt(119.25) = 10.92 seconds (rounded to two decimal places).
Now that we have the time, we can calculate the horizontal distance (d) using the formula:
d = v * t,
where
d is the horizontal distance,
v is the horizontal velocity of the Twinkie™ (2.2 m/s in this case),
and t is the time of flight we just calculated.
Plugging in the values:
d = 2.2 * 10.92 = 24.02 meters (rounded to two decimal places).
Therefore, the horizontal distance the Twinkie™ will travel before hitting the floor of the crater is approximately 24.02 meters.