# physics help?

An astronaut drops a camera from rest while exiting a spacecraft on the Moon. The camera falls 2.3 m in 1.7 s. Calculate (a) the acceleration due to gravity on the Moon and (b) the impact velocity of the camera with the surface

### 4 Answers

- 5 months agoFavorite Answer
(a) From Newton's Universal Law of Gravitation: g = GM/r, where:

*G = the gravitational constant, 6.67*10^-11 N*m^2/kg^2

*M = the mass of the object = 7.35 * 10^22 kg

*r = radius of the object = 1.74 * 10^6 m

Plugging these values into the formula gives us g = 1.63 m/s^2

Alternatively (and in retrospect probably more what you wanted) you can use kinematics:

x = 1/2 * a * t^2 + v0 * t

2.3 m = 1/2 * a * (1.7 s^2) + 0 (initial velocity is zero because it's from rest)

a = 1.59 m/s^2

The answers are slightly different, but I'm going to bet that's because of round-off error.

(b) The distance dropped is very small compared to the radius, so acceleration can be assumed constant.

v = v0 + a * t

v = (0) + 1.59 m/s^2 * 1.7 s = 2.71 m/s

Hope this helps!

- oubaasLv 75 months ago
2h = g*t^2

g =2h/t^2 = 4.6/1.7^2 = 4.6/2.89 = 1.6 m/sec^2

V = √ 2gh = √ 3.2+2.3 = 2.71 m/sec

- SpacemanLv 75 months ago
(a) the acceleration due to gravity on the Moon

d = distance through which the camera falls = 2.3 m

t = elapsed time of the fall = 1.7 s

a = gravitational acceleration on the Moon = to be determined

t = √(2d / a)

t² = 2d / a

t²·a = 2d

a = 2d / t²

a = 2(2.3 m) / (1.7 s)²

a = 1.591695502 ms² = 1.6 m/s²

(b) the impact velocity of the camera with the surface

vi = instantaneous velocity of an object after falling for t seconds = to be determined

vi = a·t

vi = 1.591696 m/s²·1.7 s

vi = 2.705882353 m/s 2.7 m/s

- 5 months ago
a(t) = a

v(t) = a * t + C

v(0) = 0

v(t) = a * t

s(t) = (a/2) * t^2 + C

s(0) = 2.3

s(t) = (a/2) * t^2 + 2.3

s(1.7) = 0

0 = (a/2) * 1.7^2 + 2.3

-2.3 = 2.89 * a / 2

-4.6 = 2.89 * a

a = -4.6 / 2.89

a = -460 / 289

v(1.7) = (-460/289) * 1.7 = (-460 / 17^2) * (17/10) = -46/17