Motion Q (1 Viewer)

[Teoh]

Just totally forgotten the process of how to work through this

A man stands on the roof of a tower and shoots an arrow in a horizontal direction. The arrow starts 45m above the ground with an initial speed of 30m/s. Assume G=10m/s/s

What time elapses before the arrow strikes the ground?

How far from the base of the tower does it hit the arrow?

What is the acute angle @ between the path of the arrow, and point of impact?

EDIT: I'm guessing if I get the first part (an equation) I could probably do the next 2...*hopefully* )

 

[nike33]

this isnt really a thinking question, derive the forumulas, sub values in and ur set !

 

[Teoh]

Yeah I know...I just forgot how to derive the formulas

ITs basically
y'' = -10
so
y' = -10t + C?

IS that anywhere near right?

 

[CM_Tutor]

That's the correct start. Now, at t = 0, y' = 0 (as fired horizontally), and integrating again, you find y = -5t² + 45.

You then need to work on the horizontal components. Starting from x'' = 0, you should get x = 30t.

Remember with the last part, about the angle, you need to look at velocity equations.

 

[Teoh]

I think it's a little worrying, if I just blatently forget the process...

Hmm, and that's where I went wrong again:

Why is x'= 30t, and not just 30...

Although now the questoin does seem a lot easier...
:mad1:


EDIT: Correct assumption to say
y''/x'' = c
y'/x' = bt
y/x = at^2


???

 

[CM_Tutor]

Originally posted by Teoh
Why is x'= 30t, and not just 30...

Look at my post again. I said x'' = 0 leads to x = 30t, not to x' = 30t. You should get x' = 30 as an intermediate step

As for the last bit of your post, I'm confused

Are you trying to say that if y = at²
then dy/dt = 2at
and d²y/dt² = 2a?

 

[Teoh]

Erm... just disreguard that comment

All is good now thanks for the help CM_Tutor