# Mechanics help

Hey for this q how do I know the ratio of gravity part I understand the velocity ratio but not the g
(edited 12 months ago)
The distance theyre talking about is the horizontal distance and this is proportional to the time of flight as the horizontal velocity is constant, so the times of flight has the same ratio. So how are the times and acceleration related when the initial velocity is the same? v = u + at should be the key suvat.
(edited 12 months ago)
Original post by mqb2766
The distance theyre talking about is the horizontal distance and this is proportional to the time of flight as the horizontal velocity is constant, so the times of flight has the same ratio. So how are the times and acceleration related when the initial velocity is the same? v = u + at should be the key suvat.

I don’t understand sorry plz explain 😭
It would help (you) to be more specific about what you dont understand. How do you imagine the horizontal and vertical velocities impact on the problem? What do you sketch, equations, ... But to cut to the chase, youd get for the vertical motion
at = constant
so the acceleration ratio (gravity) is the inverse of the time (horizontal distance) ratio.
Original post by Alevelhelp.1
Hey for this q how do I know the ratio of gravity part I understand the velocity ratio but not the g

Good to see that you “exercise common sense” to see that you have not done things correctly.

In this question, the 50 m and 300 m are the respective ranges of projectile motion of the golf ball.

If you don’t recall what is the range in projectile motion have a look at this link.
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Physics_(Boundless)/3%3A_Two-Dimensional_Kinematics/3.3%3A_Projectile_Motion

The range depends on the acceleration due to gravity on Earth and the moon.
Original post by Eimmanuel
Good to see that you “exercise common sense” to see that you have not done things correctly.

In this question, the 50 m and 300 m are the respective ranges of projectile motion of the golf ball.

If you don’t recall what is the range in projectile motion have a look at this link.
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Physics_(Boundless)/3%3A_Two-Dimensional_Kinematics/3.3%3A_Projectile_Motion

The range depends on the acceleration due to gravity on Earth and the moon.

No need to be rude… this isn’t my working it’s my teachers for your information.
Do you understand how
* Using (constant) horizontal motion, the time of flight will also be in the ratio 1:6
* Using vertical motion (under gravity) with v = u+at, at the end v = -u, so at = -2u which is constant.
* So "at" on earth equals "at" on the moon. So acceleration is in the ratio 6:1
(edited 12 months ago)
Original post by mqb2766
Do you understand how
* Using (constant) horizontal motion, the time of flight will also be in the ratio 1:6
* Using vertical motion (under gravity) with v = u+at, at the end v = -u, so at = -2u which is constant.
* So "at" on earth equals "at" on the moon. So acceleration is in the ratio 6:1

I get the first part but for the second part can I tell me what to sub in the suvat formula I’m so dumb I know sorry
The key thing to understand is the given info is horizontal motion (distance ratio) but theyre asking about vertical motion (acceleration ratio). Time of flight links the two. The time of flight is in the ratio 6:1

If you throw a ball up (vertical motion on the earth/moon/...) it returns with the same speed but opposite direction, so v = -u and using the v=u+at suvat for vertical motion,
at = -2u
The right hand side is constant (earth/moon/...) as the question says the initial velocity is the same. So at on the earth equals at on the moon. So
gt = GT
where g,t refers to the earth, G,T refers to the moon or
g/G = T/t = 6/1
which is what youre after
(edited 12 months ago)