- #1

heres my thoughts on what to do: i can use 16 degrees and 5.1 m/s to find the x and y component of the initial velocity. that's all i've got so far.

any help would be appreciated

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- Thread starter physics=headache
- Start date

- #1

heres my thoughts on what to do: i can use 16 degrees and 5.1 m/s to find the x and y component of the initial velocity. that's all i've got so far.

any help would be appreciated

- #2

gnome

- 1,037

- 1

Right. So what's the problem? Once you've done that, how long does it take him to travel 4.9m horizontally? Then, given his initial VERTICAL velocity, what is his vertical displacement after that length of time?i can use 16 degrees and 5.1 m/s to find the x and y component of the initial velocity

- #3

ok i got that, but where does the 1.9 m come in?

- #4

gnome

- 1,037

- 1

Presumably it's there to help you learn to distinguish between relevant and extraneous information.

- #5

frankR

- 91

- 0

Originally posted by gnome

Right. So what's the problem? Once you've done that, how long does it take him to travel 4.9m horizontally? Then, given his initial VERTICAL velocity, what is his vertical displacement after that length of time?

y(t) = -9.81*t^2 + Voy*t + 1.9m

The jumpers initial height is 1.9m.

- #6

enigma

Staff Emeritus

Science Advisor

Gold Member

- 1,757

- 17

the height you need to find is with respect to the taller building.

The 1.9m is there because either:

1) The jump isn't enough to clear the gap.

or

2) It's there to trip you up with extraneous information.

- #7

gnome

- 1,037

- 1

When I fire from the hip, I tend to shoot myself in the foot.Presumably it's there to help you learn to distinguish between relevant and extraneous information.

Working out the actual numbers, it seems that when he reaches the second building, his vertical displacement

is -3.49m; i.e. his feet hit the wall 1.59m (3.49-1.9) below the roof. That's where the 1.9m height difference comes in.

- #8

hhegab

- 42

- 0

hhegab

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