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?

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

1.63 m/s^2 is the correct 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)

Well, that's an interesting question, friend! It seems like Neil Armstrong was having a bit of fun on the moon with his moon jumps. Now, to determine the gravitational acceleration he experienced, we can use a little physics!

First, let's think about the situation. When Neil jumped, he reached a height of 0.700m. Now, we know that the initial speed of his jump was 1.51 m/s.

Using the equation for the conservation of energy, we have:

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

Now, let's break it down:

0.5 * m * (1.51 m/s)^2 + m * g * h = 0.5 * m * (0 m/s)^2 + m * g * 0.700 m

The mass, m, cancels out on both sides of the equation. And since 0^2 is 0, we can simplify further:

0.5 * (1.51 m/s)^2 + g * 0.700 m = g * 0

If we solve for g, we find that the gravitational acceleration on the moon that Neil experienced was...wait for it...0 m/s²!

Yep, that's right! In the low-gravity environment of the moon, Neil essentially experienced weightlessness during his jump. Talk about taking a giant leap for a small gravity, right?

To determine the gravitational acceleration experienced by Neil Armstrong on the moon, we can use the equations of motion. We'll start with the equation for calculating the maximum height reached in a projectile motion:

v^2 = u^2 + 2as

Where:
v is the final velocity (0 m/s at the highest point for the jump),
u is the initial velocity (1.51 m/s),
a is the acceleration due to gravity,
s is the displacement (0.7 m).

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)

Substituting the given values, we get:

a = (0^2 - 1.51^2) / (2 * 0.7)

Simplifying the equation further:

a = (-1.51^2) / (1.4)

Calculating the value:

a = -2.557 m/s^2

Therefore, Neil Armstrong experienced a gravitational acceleration of approximately -2.557 m/s^2 on the moon. The negative sign indicates that the acceleration is directed towards the surface of the moon.

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.

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