Solving for Horizontal Distance and Time in a Vertical Drop Problem

[chanv1]

Homework Statement

A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 12.0 m/s, and the distance from the limb to the level of the saddle is 2.00 m.

What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?

____ m

(b) How long is he in the air?

____ s

Homework Equations

x(t) = x_0 + v_0(t) + (1/2)(a)(t)^2

The Attempt at a Solution

For (a) I've tried to plug numbers into the equation above

x(t) = 0 + 12(.639) + (1/2)(-9.8)(.639)^2
x= 5.67 ... but that was incorrect

for (b) I used the equation (1/2)(a)(t)^2 and got t = .639 s

I don't know what I'm doing wrong for part (a), can someone please teach me? Thanks.

 

[CompuChip]

Personally, I would use the result of b) to answer a).
How much distance does the horse cover while the ranch hand is in the air?

 

[merryjman]

Compuchip is right - use the time to find out the answer to (a). The problem with your answer to (a) is that you've got a horizontal velocity (12 m/s) and a vertical acceleration (9.8 m/s/s) in the same equation. That's not allowed. Velocities and accelerations are vectors, so X stuff can only affect other X stuff, and Y stuff can only affect other Y stuff.

But now that you know how long the ranch hand is in the air, you can figure out how far the horse moves in that time. Use the same "relevant equation" you listed, and use 12 m/s for the initial velocity, but think carefully about the horse's acceleration. It's not 9.8 m/s/s.

 

[DaveC426913]

But now that you know how long the ranch hand is in the air, you can figure out how far the horse moves in that time. Use the same "relevant equation" you listed, and use 12 m/s for the initial velocity,

No. The horse is moving at a constant speed as given. The formula giving distance from time and speed is ... quite a bit simpler...

 

[merryjman]

Right, Dave, so if you use the equation I said, and put in the correct acceleration, you get the simple formula you're talking about, don't you, bro??!?

 

[CompuChip]

No. The horse is moving at a constant speed as given. The formula giving distance from time and speed is ... quite a bit simpler...

Exactly merryjman.
Indeed, as the topic poster quoted the most general formula for such cases, I encourage him to identify a, v0 and x0 in each situation and apply it, rather than learning all the different formulas which are applicable in different cases, and can be derived from the general one.

 

[DaveC426913]

Uh, technically true, but aren't you sort of making extra work for the OP? I mean the formula for d as a function v and t is about as basic as you can get. You seriously think it's better for him to use the acceleration formula for this?

 

[merryjman]

I definitely see your point, Dave, but in my experience it's been valuable to show that the acceleration formula reduces to the simpler version. It helps reinforce the vector nature of velocity and acceleration through repetition, and like Compuchip said, it discourages students from memorizing shortcut formulas.