problem A newly established colony on the Moon launches a capsule vertically with an initial speed of 1.445 km/s. Ignoring the rotation of the Moon, what is the maximum height reached by the capsule

Answer:

104.4km

Explanation:

Projectile motion occurs when object is launched into air and allowed to fall freely under the influence of gravity.

Maximum height reached by the object is expressed as;

H = u²sin²(theta)/2g where;

u is the initial velocity = 1.445km/s

u = 1445m/s since 1000m is equivalent to 1km

theta is the angle of launch

g is the acceleration due to gravity = 10m/s²

theta = 90° (since object is launched vertically)

Substituting the values in the formula we have;

H = 1445²(sin90°)²/2(10)

H = 1445²/20

H = 104,401.25m

H = 104.4km

The the maximum height reached by the capsule is 104.4km

Answer:

104.4km

Explanation:

Projectile motion occurs when object is launched into air and allowed to fall freely under the influence of gravity.

Maximum height reached by the object is expressed as;

H = u²sin²(theta)/2g where;

u is the initial velocity = 1.445km/s

u = 1445m/s since 1000m is equivalent to 1km

theta is the angle of launch

g is the acceleration due to gravity = 10m/s²

theta = 90° (since object is launched vertically)

Substituting the values in the formula we have;

H = 1445²(sin90°)²/2(10)

H = 1445²/20

H = 104,401.25m

H = 104.4km

The the maximum height reached by the capsule is 104.4km

Answer:

Maximum height =1031km

Explanation:

Given:

Velocity,Vo= 1.445km/s= 1445m/s

Mass of moon= 7.35×10^22kg

Radius of moon= 1737km= 1737000m

Using conservation energy

Ui + Ki= Uf + Kf

–‘>(GMm/R) + 1/2 (m ^2)

-GMm/(R + h) – 0

Vo^2= 2Gm(1/R – 1/R+h)

1445^2= 2× (6.67×10^-11)×(7.35×10^22)[(1/1737000)- (1/1737000 + h)]

h= 1031km