Solving for Height in Horizontal Projectile Motion: 41 m/s, 23 m, g = -9.8m/s

[NemoMnemosyne]

This was an even numbered problem from the book so there are no answers to make sure I've done this correctly or not.

Homework Statement

An arrow fires horizontally at 41 m/s travels 23 m horizontally. From what height was it fired?

Vi = 41 m/s
a = g = -9.8m/s
x = 23m

Homework Equations

Vf = Vi + at
X = Xi + Vit + 1/2at^2
Y = Yi + Vit + 1/2at^2

The Attempt at a Solution

A.) Time to reach ground

Vf = Vi +at =>
Vf - Vi/a = t =>
0 - 41/-9.8 = 4.2s

B.) Height of Cliff
y = Vit + 1/2at^2 =>
y = 41(4.2) + 1/2(-9.8)(4.2^2) =>
y = 85.76m

 

[davo789]

Hi!
It appears you have got yourself into a bit of a muddle. The real trick with questions like this is to separate out all the x and y parameters (speed (initial, u, and final, v), acceleration (a), distance (s), time (t)).

So, in x, we have:

Vx = Ux = 41m/s
ax = 0 (nothing to accelerate the particle in the x plane)
sx = 23m
tx = ?

And in y,

Uy = 0
Vy = ?
ay = g = 9.8m/s/s
sy = what we want to find
ty = tx (can you see why?)

Now, clearly you want to know how long the arrow is in the air for. You haven't got enough information from the y variables, but you have plenty in x! It turns out you just need to do a simple speed = distance over time calculation.

See what you can do with this new info.

 

[Spinnor]

You have the right set of equations but you are getting x and y directions mixed up. The arrow is fired horizontally and that horizontal velocity is unchanging. The initial velocity in the vertical direction is zero and changes constantly until impact.