Net displacement with 3 figures

[ikihi]

Homework Statement

A ball on a porch rolls 60 cm to the porch's edge, drops 40 cm, continues rolling on the grass, and eventually stops 80 cm from the porch's edge.

What is the magnitude of the ball's net displacement, in centimeters?

Homework Equations

d = a^2 + b^2

The Attempt at a Solution

So there seems to be two triangles, each having 2 known sides. So I found the displacement of sides 60 and 40 and got 72.111 cm
Then for the 2nd triangle I found the displacement of sides 40 and 80 and got 89.443 cm.

The next step I was confused on. But I went ahead and put the two vectors together and I used the p.theorem to take the square root of 89.443^2 and 72.111^2 and came up with 114 cm as the net displacement. Is this right or way off. My dumb book doesn't give the answer.

 

[haruspex]

the square root of 89.443^2 and 72.111^2 and came up with 114 cm as the net displacement.

Pythagoras' Theorem is for a right-angled triangle. Are your 89.443 and 72.111 two sides of a right-angled triangle?
Did you draw a diagram of the whole process?

 

[nil1996]

Homework Equations

d = a^2 + b^2

should it be
d²=a²+b²

 

[ikihi]

Pythagoras' Theorem is for a right-angled triangle. Are your 89.443 and 72.111 two sides of a right-angled triangle?
Did you draw a diagram of the whole process?

72 cm and 89 cm are the hypotenuse's of two different triangles.

The first triangle is 40 and 60; The hypotenuse is 72 cm. The second is 40 and 80: this one's hypotenuse is 89 cm. I need to find the magnitude of the ball's net displacement.

ball starts rolling here
\/
60cm->
#----------#
******** -
******** - <-40 cm
******** -
******** #-----------------# <---ball stops rolling here

......80cm ^

 

[ikihi]

should it be
d²=a²+b²

That is the theorem, however if the length of both a and b are known, then c, the hypotenuse, can
......... _________
be calculated as follows: c=√a^2+b^2.

 

[haruspex]

72 cm and 89 cm are the hypotenuse's of two different triangles.

The first triangle is 40 and 60; The hypotenuse is 72 cm. The second is 40 and 80: this one's hypotenuse is 89 cm. I need to find the magnitude of the ball's net displacement.

ball starts rolling here
\/
60cm->
#----------#
******** -
******** - <-40 cm
******** -
******** #-----------------# <---ball stops rolling here

......80cm ^

You are assuming it is useful to calculate those two hypotenuses, but if you draw in your diagram where those two distances are you'll see that they will not help you. They don't even join up. You need to use a little geometry. Let's say it starts at A, rolls to B, drops (vertically) to C, then continues to D. You want AD. Try projecting AB horizontally to E directly above D and joining DE.

 

[ikihi]

You are assuming it is useful to calculate those two hypotenuses, but if you draw in your diagram where those two distances are you'll see that they will not help you. They don't even join up. You need to use a little geometry. Let's say it starts at A, rolls to B, drops (vertically) to C, then continues to D. You want AD. Try projecting AB horizontally to E directly above D and joining DE.

So we would add the total x-axis 60+80 and get 140?
Then the y-axis of the triangle is 40.So it would be: d= √(140^2+40^2)

that would equal 145.6 cm. Since it has 2 sig figs does that mean it is ≈146 cm?

 

[haruspex]

...... ____________
So it would be: d= √140^2+40^2

Yes. (How about using √( ), it would be much easier to type.)

that would equal 145.6 cm. Since it has 2 sig figs does that mean it is ≈146 cm?

2 sig figs would make it 150. Are you sure that's what the question is asking for?