Question

Help please!
Free fall question:
A body freefalls from a tower that is 25m. Find the displacement equation using integral calculus. Note: acceleration due to gravity =-9.8m/s^-2

Answer 1

OK - this is quite a standard task.

acceleration = dv/dt and in this case is a constant (g)

so first you have
\[a=\frac{ dv }{ dt }\]

so

\[v= \int\limits a dt\]

do this integral first - a = constant = g.

Don't forget the constant of integration

Answer 2

We can start with the equation that relates the acceleration to the time rate of change of speed as @MrNood explained above.

Answer 3

Once you have done the integral you can calculate the value for the constant of integration based on the fact that the body 'falls' therefore it started with velocity = 0 when t=0

Answer 4

v = at

Answer 5

have you include the constant of ingration and worked it out? (no!)

Answer 6

sorry, v=at +c

Answer 7

ok - to get the complete equation you have to say that at time t=0 the velocity is u

(in THIS case u=0- but I suggest you work with a value u for now)

so work out the value of c given t=0 v=u

Answer 8

u = a*0+c
u = c
Therefore, v=at+u

Answer 9

OK good - maybe you recognise that equation?

now v = u+at = ds/dt

so
\[s= \int\limits_{}^{} v dt = \int\limits_{}^{}u +at dt\]

do this integral

AND - don't forget c

Answer 10

It is
\[ut+\frac{ at ^{2} }{ 2 } +c\]

Answer 11

good - that is mainly your answer.

but you need to define your frame of reference and datum.

we to a as positive - i.e. the direction of acceleration is positive downwards

Now we need to set a datum. The tower is 25 m high.

IF we set the initial position = 0 then ground level is +25m

IF we set the ground level as 0 then initial position is -25m

this decision is not given in the question - but affects the value of c in th second equation.

Hope that is clear

Well done

Answer 12

yup, thats clear

Answer 13

does it give the same answer?

Answer 14

No

in one case c= 0, in the other c= -25

in all cases the valu of c in an integration depends on some known conditions.

In the case of motion equations it is 'normal' (but not compulsory) to take the displacement s=0 when t= 0

Just be aware of the sign you use for a, v &s = they must all have the same direction

Answer 15

Okay. :)

Answer 16

note - that doesn't mean they all have the same sign!

if you measure displacement and speed upwards (say a ball thrown upwards at speed u) then acceleration has NEGATIVE sign as it is in ht eopposite direction to measuring speed and height......

Answer 17

so, when a ball is dropped, the acceleration is positive and the velocity and displacement are negative.

Answer 18

?

Answer 19

No - I don't mean to confuse you.

The main thing is that there is no 'Absolute' reference. We CHOOSE the reference when we set up an equation.

In your particular question we chose to set a= +ve downmards, and then we must take velocity and dispalcemet as positive downwards.

In another question for instance it may say ' a body ois projected up at 5m/s - find maximum height.
in THAT case we implicitly take speed and displacement as +ve Upwards BUT acceleration is therefore -ve

Answer 20

Oh, okay. i think i understand now. Thank you :)