Please check my Answer to Free Fall Question

1. Nov 28, 2008

cbrowne

1. The problem statement, all variables and given/known data

A man standing on the sidewalk throws a set of keys vertically upward to his
wife in the window of their apartment 4.0 m above. The wife catches the keys
1.5 s later. (a) With what initial speed were the keys thrown? (b) What is the
velocity of the keys just before they were caught? (c) Draw a diagram showing
the trajectory of the keys.

2. Relevant equations

3. The attempt at a solution

d= 4.0 m
Δt= 1.5 s
g= -9.8 m/s^2

Δd= ViΔt+1/2aΔt

-ViΔt= 1/2 at2-Δd

-Vi= 1/2 at2- Δd /Δt

-Vi= -10.016 m/s

ANSWER FOR 7-A ->>Vi= 10.016 m/s

Vf^2 = Vi^2 +2ad

Vf= 4.68333 m/s

2. Nov 29, 2008

alphysicist

Hi cbrowne,

I don't believe that equation will answer the question. It gives the speed, but they are asking for the velocity.

When you use the equation:

Vf^2 = Vi^2 +2ad

you have to remember that because the Vf and Vi are squared, the equation does not differentiate between positive and negative velocities. So all that equation is telling you is that after taking the square root:

$$V_f = \pm 4.68333 \mbox{ m/s}$$

(For example, what is the square root of 4? It is both +2 and -2.) So you still have to determine which sign is the correct sign. Try using another kinematic equation that has the final velocity in it to determine the direction. What do you get?