where s is the distance travelled (m)
u is the initial velocity (m/s)
a is the acceleration (m/s^2)
t is time (s)
figures in brackets represent the unit.
For your case, a is the acceleration due to gravity and it is in the opposite direction of that of the velocity.
Therefore s= ut + (0.5)(-9.81)t^2
From there you get, s= ut - 4.9t^2
done!
Callyst
4 January 2010
the formula i put down was the same except it was set to determine the height in meters. also i'm still working on this problem and i don't understand why t (seconds) is squared. i understand 9.8 becoming 4.9 but why does gravity continue to accelerate against the initial speed.
example: H= 98m/s - 4.9m/s^2
i'd like to use this for an example problem.
freakawill
8 February 2010
This explanation requires calculus.
Let's start from acceleration. 9.8m/s^2 is the acceleration due to gravity. So we can write:
a=9.8
Our y right now is a rate of change of velocity so we can write it as
da/dt=9.8
Let's integrate to find the velocity and call it v.
int(dy/dt=9.8,t)=9.8t + c
v=9.8t + c
Our c is any random arbitrary number but for this equation it is the starting velocity. But this is still the rate of change of distance so let's integrate again getting...
y=k + ct + (9.8)(1/2)t^2
or
y=k +ct + 4.9t^2
Where k is just like c but instead our initial height.
So basically we derived a general formula for an object with a given acceleration. So why are the seconds squared?
It is a rate of a rate. It is quite literally meters per second per second. To convert to meters, you must multiply by seconds twice.
dave08
18 January 2014
i know my physics teacher tell us all the formula about that and i know he teach us that formula.
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Comments (4)
the actual formula should be
s=ut + 0.5at^2
where s is the distance travelled (m)
u is the initial velocity (m/s)
a is the acceleration (m/s^2)
t is time (s)
figures in brackets represent the unit.
For your case, a is the acceleration due to gravity and it is in the opposite direction of that of the velocity.
Therefore s= ut + (0.5)(-9.81)t^2
From there you get, s= ut - 4.9t^2
done!
the formula i put down was the same except it was set to determine the height in meters. also i'm still working on this problem and i don't understand why t (seconds) is squared. i understand 9.8 becoming 4.9 but why does gravity continue to accelerate against the initial speed.
example: H= 98m/s - 4.9m/s^2
i'd like to use this for an example problem.
This explanation requires calculus.
Let's start from acceleration. 9.8m/s^2 is the acceleration due to gravity. So we can write:
a=9.8
Our y right now is a rate of change of velocity so we can write it as
da/dt=9.8
Let's integrate to find the velocity and call it v.
int(dy/dt=9.8,t)=9.8t + c
v=9.8t + c
Our c is any random arbitrary number but for this equation it is the starting velocity. But this is still the rate of change of distance so let's integrate again getting...
y=k + ct + (9.8)(1/2)t^2
or
y=k +ct + 4.9t^2
Where k is just like c but instead our initial height.
So basically we derived a general formula for an object with a given acceleration. So why are the seconds squared?
It is a rate of a rate. It is quite literally meters per second per second. To convert to meters, you must multiply by seconds twice.
i know my physics teacher tell us all the formula about that and i know he teach us that formula.
Add a Comment or Suggest an Answer