Velocity Query

Tikkal

Please assist me on this question:-

"A ball is dropped out of a window near the top of a high building. What is its velocity when it has fallen 5.0m?

Many thanks.

Spear

Tikkal wrote: »

Please assist me on this question:-

"A ball is dropped out of a window near the top of a high building. What is its velocity when it has fallen 5.0m?

Many thanks.

Homework question? Show us your workings so far.

Tikkal

I have formulas but I am unable to work it out because the only information I have is '5m' drop.

V2= U2 + 2AS

or

S = 0 + 1/2g + 2

magicianz

V squared formula-

V=v
U=0
S=5
A=-9.8
And solve

Spear

Tikkal wrote: »

I have formulas but I am unable to work it out because the only information I have is '5m' drop.

V2= U2 + 2AS

or

S = 0 + 1/2g + 2

No, you have initial velocity, initial height, acceleration and distance travelled.

Tikkal

Many thanks. Solved.

Delphi91

magicianz wrote: »

V squared formula-

V=v
U=0
S=5
A=-9.8
And solve

Surely that should be A = 9.8 since the acceleration is in the direction in which the ball is falling!

Spear

Delphi91 wrote: »

Surely that should be A = 9.8 since the acceleration is in the direction in which the ball is falling!

Matter of preference. I'd tend to use a negative number here and define the initial height as 0.

magicianz

Delphi91 wrote: »

Surely that should be A = 9.8 since the acceleration is in the direction in which the ball is falling!

Yeah it is down to preference as spear said but i prefer to keep it as -9.8 in all my questions as it reduces the amount of slips you make. (accidently changing it from 9.8 to -9.8 or vice versa)

Professor_Fink

Spear wrote: »

Matter of preference. I'd tend to use a negative number here and define the initial height as 0.

It doesn't really matter the initial height. If you use acceleration 9.8 rather than -9.8 it means you have to make displacement negitive instead, since otherwise the height increases. Which one you choose is a matter of preference, but making the acceleration negative tends to cause th least confusion.

VinnyTGM

You only gave us:
distance travelled-5m
acceleration due to gravity is 9.8ms

what about the mass of the ball??

v=u+at
s=ut+(.5)a(t)(t)
(v)(v)=(u)(u)=2as

Delphi91

VinnyTGM wrote: »

You only gave us:
distance travelled-5m
acceleration due to gravity is 9.8ms

what about the mass of the ball??

v=u+at
s=ut+(.5)a(t)(t)
(v)(v)=(u)(u)=2as

You don't need to know the mass of the ball - look at the equations you've written under your post, there is no reference to the mass (m) of the object, hence the mass is not required.

BTW, the third equation is incorrect, it should read: (v)(v) = (u)(u) + 2as