# When to use certain kinematic equations

#### lamborghiniMD

##### Full Member
Okay, so here's a problem from a practice test:

"Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time, t. If the value of g where reduced to g/6, then t would:

a) incr by 6
b) incr by 6^.5
c) decr by 6
d decr by 6^.5

Okay, so I narrowed it down to a & b, and I WAS going to pick a (the correct answer), but then of course I had to doubt myself and pick b.

I chose a at first using t = v/g. but then I switch to b, because I used h = .5at^2.

Now I'm confused. Both equations are valid for time, but why was the t = v/g correct for this problem?

#### cloak25

##### Full Member
Okay, so here's a problem from a practice test:

"Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time, t. If the value of g where reduced to g/6, then t would:

a) incr by 6
b) incr by 6^.5
c) decr by 6
d decr by 6^.5

Okay, so I narrowed it down to a & b, and I WAS going to pick a (the correct answer), but then of course I had to doubt myself and pick b.

I chose a at first using t = v/g. but then I switch to b, because I used h = .5at^2.

Now I'm confused. Both equations are valid for time, but why was the t = v/g correct for this problem?

I think you use h = .5at^2 when you assume initial velocity is 0. In this case, the ball is thrown up so it has an initial velocity, therefore you cannot use h=0.5at^2.

Instead you can use d= vot + 0.5at^2 where d = 0 because the ball returns to its original height (0 displacement). So 0 = vot + 0.5at^2

since acceleration is negative, vot = 0.5at^2. time cancels and you get t = 2vo/g, which is the equation they used to get the answer.

#### MangoPlant

##### Full Member
You cannot use h = (1/2)at² because there are tree variables in that equation that change. When gravity is reduced by a factor of 6, h and t change (increases). So because tree variables are changing you cannot use that equation. In t = v/g only two variables, g and t, change.

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#### flashlightpen

##### Full Member
Okay, so here's a problem from a practice test:

"Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time, t. If the value of g where reduced to g/6, then t would:

a) incr by 6
b) incr by 6^.5
c) decr by 6
d decr by 6^.5

Okay, so I narrowed it down to a & b, and I WAS going to pick a (the correct answer), but then of course I had to doubt myself and pick b.

I chose a at first using t = v/g. but then I switch to b, because I used h = .5at^2.

Now I'm confused. Both equations are valid for time, but why was the t = v/g correct for this problem?

Hey man!

Total off topic, but a pretty cool tip I got Whenever a question mentions time, use kinematics. If they dont want time, just use energy. Sounds kind of simple, but its been really helpful for me

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