Question

A ball is thrown from the surface of the moon with a velocity of 19.6 m/s. How much time will it take to attain maximum height? How high will it go?

Answer 1

The gravitational acceleration on the moon is 1/6 of the earth's gravitational acceleration.
$g\left(atthemoon\right)=9.8\times \frac{1}{6}=1.63m/s^2$.
Given
u = 19.6 m/s
At maximum height velocity of the object will be zero, (v=0).
using v = u + gt
Here the acceleration is -ve(as the body is decelerated), and velocity toward up is taken +ve.
0 = 19.6 + (-1.63*t)
1.63*t=19.6
t = 19.6/1.63
t = 12.02 sec.

Using the third equation of the motion,
$v^2-u^2=2gs$
$s=(v^2-u^2)/2g$
$s=(0-384.16)/2(-1.63)$
s = 117.84 m.