while on the moon, the Apollo astronauts enjoyed the effect of a gravity much smaller than that on earth. if Neil Armstrong jumped up on the moon with an initial speed of 1.51 m/s to a height of 0.700m, what amount of gravitational acceleration did he experience?

Question

drwls · Accepted Answer

Using conservation of energy, (1/2)M Vo^2 = M g'H, so g' = Vo^2/(2H) is the value of the acceleration of gravity on the moon. I get 1.63 m/s^2

Marko · Answer

2gh=v^2 2*0.7*g=1.51^2 1.4g=1.51^2 /g 1.4=2.2801g g=1.4/2.2801 g=0.6140081575 m/s^2 1.63m/s^2 is not the correct answer, take for example this: 10=5g What is 'g'? g=10/5=2

Tyandra · Answer

1.63 m/s^2 is the correct answer

Matt · Answer

Marko 1.51^2 /g doesn't equal 2.2801g it equals 2.2801/g. So your answer is the reciprical of the right answer. Nice try though.

SIXLOOP · Answer

While on the moon, the Apollo astronauts enjoyed the effects of a small gravity. If Neil Armstrong jumped up on the moon with an initial speed of 5 m/s to a height of 5.0 m, what amount of gravitational acceleration did he experience? (m/s^2)