# Projectile motion

1. Sep 24, 2011

### ibysaiyan

1. The problem statement, all variables and given/known data
The question is as following:
a) A dog in a stunt takes a jump off a rook which is 40 m high above the ground and lands on the next building which is of the same height , the gap between the two buildings is 6m @ the velocity of 8 m/s .Find the range of angles at which he must take off.

There are more questions which I will post on follow up.

2. Relevant equations

3. The attempt at a solution

Now the first bit I thought of sketching a triangle in correspondence to the information available... what I found out is that we have two distances given so to find angle I would use tan theta = opposite/Adjacent = 40/6 which gives me an angle of about 81. I am also aware of the general solution of tan... which's npi +180.. however don't these range of angles look too absurd,large ?

b)

2. Sep 24, 2011

### PeterO

Firstly, did the dog really jump from a rook or did you mean roof?

Secondly, Since both buildings were 40m high, the problem would e the same if they were in fact 2m high [and a lot safer], so I wouldn't go using the 40m in a trig function the way you did.
What is required is that the range of this jump has to be 6m.
While you could start from first principles, and resolve the velocity into a vertical and horizontal component in order to get flight time etc, you may have already derived an equation for the range of a projectile that lands at the same height as it was launched from [usually used to find out how far away something lands on flat ground]. If you have that equation this problem is very simple to solve.
Try looking in your text under range - or even google range of a projectile.

3. Sep 24, 2011

### ibysaiyan

Yes,I meant to type roof (typo). I am still unsure on how to find range of angles...

4. Sep 24, 2011

### PeterO

http://library.thinkquest.org/29263/rangee.htm

The above page shows the derivation of the formula for the range of a projectile, in the situation where the landing height is the same as the launch height.

The final formula includes an angle.

You want the range to be greater than or equal to 6m [it is OK to land a little further onto the roof; you just must not land short]
The nature of projectiles is:

project at 45o for maximum range.

43o and 47o give the same range as does any other pair of angles equally each side of 45o.

You may find a range of 6m for 39o, in which case 51o would also work, so the range of angles would be from 39o to 51o.

NOTE: I just made up 39o as a possibility for explanation purposes. If it turns out to be correct it was a lucky guess!!

Last edited by a moderator: Apr 26, 2017
5. Sep 24, 2011

### ibysaiyan

Thanks for your reply Petero. The angle which I managed to get is 33.5 degree at the range of 6m. Your post leads me to the following question.. as you said 51 could also be the possible angle ,in this there'd be many angles at which the dog can jump off the edge safely, however how do I find out the critical angle beyond which it'd be unsafe. Obviously any angle below 33.5 would give a range of >6m...

EDIT : I just re-read your post.... would the angle range be 33.44 < theta < 45 ( since 45 is the max. range ? )

Last edited by a moderator: Apr 26, 2017
6. Sep 24, 2011

### PeterO

If 33.44 is indeed the small angle answer, the large angle limit is 56.56

33.44 < theta < 56.56

Those two angles are equally above and below 45 degrees.

7. Sep 24, 2011

### ibysaiyan

Oh right.. so by subtracting whatever angle we get from 45 would give us the other angle which is also equally sized to 45 degree.

for e.g: If I had got an angle of say 24 , then the other critical value would be 66.

Have I understood this right ?

8. Sep 24, 2011

### PeterO

That is correct in calculation.

It comes about as the Range has a Sin(2A) term in the expression.

The sine function is symmetrical about 90 degrees, eg sin50 and sine 130 have the same value.

So when we halve it [because it was sin(2A)] we get two answers symmetrical around 45.

9. Sep 25, 2011

### ibysaiyan

Thank you very much! This all makes sense now.