Finding the max range of a jump on the moon using kinematics

[garcia1]

Homework Statement

A person can jump a maximum horizontal
distance (by using a 45 ◦
projectile angle) of
4 m on Earth.
The acceleration of gravity is 9.8 m/s
2
.
What would be his maximum range on the
Moon, where the free-fall acceleration is g
6 ?
Answer in units of m.

Homework Equations

Kinematics equations

The Attempt at a Solution

All I can think to do is set kinematics equations for the x and y-axis equal to each other by another variable, such as time. This is as far as I can figure out with this problem.

 

[tiny-tim]

hi garcia1!

(have a degree: ° and try using the X² icon just above the Reply box )

from the first part, find the speed at which this person can jump

then use that speed with the Moon's g to find how far he will jump on the Moon …

what do you get? ​

 

[garcia1]

24.0304m/s, by using the fact that vf = 0, and then solving for VoY and getting 6.26m/s. I then plugged this into:

Y = Vy^2 - VoY^2 / (2*-9.81m/s^2 / 6) = 12.01518

Multiplying by two because I used Vf = 0 at the top of the jump, I got 24.0304m. It was right!

 

[tiny-tim]

Woohoo! ​

ok, now have you noticed that the range is inversely proportional to the gravity (24/4 = 6)?

can you prove that, and so avoid all the tedious arithmetic? ​

 

[rwisz]

You are on the right track with using kinematics equations to solve this problem. To find the maximum range of a jump on the moon, we can use the equation for horizontal displacement:

d = v0x * t

Where d is the horizontal displacement, v0x is the initial horizontal velocity (which can be found using the given projectile angle of 45 degrees), and t is the time of flight.

To find the time of flight, we can use the equation for vertical displacement:

h = v0y * t - 1/2 * g * t^2

Where h is the maximum height reached during the jump, v0y is the initial vertical velocity (which can be found using the given projectile angle of 45 degrees and the free-fall acceleration of the moon, g = 1.6 m/s^2), and t is the time of flight.

We can rearrange this equation to solve for t:

t = (v0y ± √(v0y^2 + 2gh)) / g

Since we want to find the maximum range, we will use the positive solution for t.

Substituting this value of t into the equation for horizontal displacement, we get:

d = v0x * [(v0y + √(v0y^2 + 2gh)) / g]

Now we can plug in the values for v0x and v0y and solve for d:

d = (v0 * cos(45)) * [(v0 * sin(45) + √((v0 * sin(45))^2 + 2 * 1.6 * h)) / 1.6]

Simplifying this equation, we get:

d = 1.25 * [(1.25 + √(1.5625 + 3.2 * h))]

To find the maximum range, we need to find the value of h that will result in the largest possible value for d. This occurs when the term inside the square root is equal to 0, which means that the maximum height reached during the jump is 0. This makes sense, as the maximum range will occur when the person jumps at a 45 degree angle and lands at the same height they started at.

Therefore, the maximum range on the moon would be:

d = 1.25 * [(1.25 + √(1.5625 +