A projectile is launched vertically from the surface of the Moon with an initial speed of 1210 m/s. At what altitude is the projectile's speed two-fifths its initial value?
Oh well, as a first guess since you did not say I will assume g moon = g earth/6
= 9.8/6 = 1.63 m/s^2
so
a = -1.6
v = Vi - 1.6 t
(2/5)(1210) = 1210 - 1.6 t
1.6 t = 1210 (3/5)
t = 454 seconds
h = Vi t - ((1/2)(1.6) t^2
= 384,447 meters
the mass of the moon is M=7.35e22
and the radius= 1.738 e6
I used the formula
1/2m(2/5vi)^2 - GMm/h+R= 1/2vi^2-GMm/R
Where little m cancels out and solved for h. But my answer and the answer you just gave me was marked wrong.
So you have any other suggestions?
well, what did you calculate for g moon? If it is not something like 9.81/6 it is wrong, but the 1/6 of earth gravity is a rough approximation.
6.67e-11
Your way I get
(1/2)(21/25) vi^2 = G M [1/R - 1/(R+h) ]
21/50 (1210)^2 = 6.67*10^-11 * 7.35*10^22
[ 1/1.74*10^6 - 1/(1.74*10^6+h) ]
.615*10^6=49*10^11 (.575*10-6 - z)
where
z = 1/(1.74*10^6+h)
.1255*10^-6 = .575*10^-6 - z
z = .449*10^-6
1/z = 2.22*10^6 = 1.74 *10^6 + h
h = .485 *10^6
485,000 meters
check our arithmetics
using your figures
g moon = F/m = G M/r^2
= 6.67^10^-11*7.35*10^22 /3.03*10*12
=16.17*10^-1 = 1.67 m/s^2
I guessed 9.8/6 = 1.63 so I would be a little off
okay, I got it thanks !
Great !
To find the altitude at which the projectile's speed is two-fifths its initial value, we need to apply the laws of motion, particularly the equations of motion for a vertically launched projectile.
First, let's analyze the given information:
- Initial speed (u) = 1210 m/s
- Final speed (v) = (2/5) * initial speed = (2/5) * 1210 m/s
- Acceleration due to gravity (g) = 1.62 m/s² (on the Moon)
Now, let's use the equations of motion to find the altitude (h) at which the speed is two-fifths its initial value:
1. The equation for the final speed (v) in terms of initial velocity (u), acceleration (g), and height (h) is given by:
v² = u² + 2gh
Rearranging the equation, we can solve for h:
h = (v² - u²) / (2g)
2. Substituting the given values into the equation:
h = [(2/5) * 1210 m/s]² - (1210 m/s)² / (2 * 1.62 m/s²)
Calculating this equation will yield the desired altitude.