Please check my Answer to Free Fall Question

  • Thread starter Thread starter cbrowne
  • Start date Start date
  • Tags Tags
    Fall Free fall
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
cbrowne
Messages
30
Reaction score
0

Homework Statement



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.

Homework Equations


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= √(10.016m/s)2+2ad

Vf= 4.68333 m/s
 
Physics news on Phys.org
Hi cbrowne,

cbrowne said:

Homework Statement



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.

Homework Equations





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= √(10.016m/s)2+2ad

Vf= 4.68333 m/s

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:

[tex] V_f = \pm 4.68333 \mbox{ m/s}[/tex]

(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?