Question

Suppose an object is thrown with the same initial velocity and direction on Earth and on the Moon, where the acceleration due to gravity is one-sixth its value on Earth. How will vertical velocity, time of fight, maximum height, and horizontal distance change?
I would appreciate if they were numerical values like doubles or triples. Thank you

Answer 1

As we know that

Time of flight is given by

$T = \frac{2vsin\theta}{g}$

Vertical height

$H = \frac{v^2sin^2\theta}{2g}$

Horizontal range

$R = \frac{v^2sin2\theta}{g}$

vertical speed

$v_y = vsin\theta$

now we can see all factors except vertical speed depends on gravity inversely

So here we can say that

Range on moon will increase by 6 times

Maximum vertical height increased by 6 times

Time of flight increased by 6 times

as all above depends inversely on gravity

vertical speed remains unchanged as it is independent of gravity

Answer 2

Let's call

g = acceleration on earth

g '= acceleration on the moon.

We know that:

$g'= \frac{1}{6}g$

1. The formula for vertical velocity on earth is:

$V_y = V_0sin(\theta) - gt$

On the moon, we have:

$V_y'= V_0sin(\theta) - \frac{1}{6}gt$

$V_y -V_y'= V_0sin(\theta) - gt - V_0sin(\theta) + \frac{1}{6}gt$

$Vy -Vy'= \frac{1}{6}gt -gt$

$Vy -Vy'= gt(\frac{1}{6} -1)$

$Vy'= Vy + gt(\frac{5}{6})$

The vertical speed on the moon is greater than on earth by a factor of gt (5/6)

2. The formula for the time of flight on earth is:

$t_v = 2\frac{V_0sin(\theta)}{g}$

On the moon it is:

$t_v'= 2\frac{V_0sin(\theta)}{(\frac{1}{6}g)}$

$t_v'= 12\frac{V_0sin(\theta)}{g}$

$\frac{t_v'}{t_v} = \frac{12}{2}$

$t_v'= 6t_v$

The time of flight on the moon is 6 times greater than on earth.

3. The maximum height on earth is:

$h = \frac{V_0^2sin^2(\theta)}{2g}$

On the moon:

$h' = \frac{V_0^2sin^2(\theta)}{2(\frac{1}{6}g)}\\\\h' = 3\frac{V_0^2sin^2(\theta)}{g}$

So:

$\frac{h'}{h} = \frac{[3\frac{V_0^2sin^2(\theta)}{g}]}{[\frac{V_0^2sin^2(\theta)}{2g}]}$

$h'= 6h$

The maximum height on the moon is 6 times greater than on earth.

4. On earth, the horizontal distance traveled is equal to:

$r_x = V_0cos(\theta)t_v\\\\r_x'= V_0cos(\theta)(6t_v)\\\\\frac{r_x'}{r_x} = 6\\\\r_x'= 6r_x$

The horizontal distance traveled on the moon is 6 times greater than that of the earth.