One Dimensional Motion Question

  • Thread starter Bahadar
  • Start date
  • #1
Bahadar
4
0

Homework Statement


Pelicans tuck their wings and free-fall straight down when diving for fish. Suppose a pelican starts its dive from a height of 18.0m and cannot change its path once committed. If it takes a fish 0.25s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fish is at the surface of the water. (Answer must be in 2 significant figures.)


Homework Equations


X=vot+1/2at^2


The Attempt at a Solution


I tried to go about solving this but it seems im still getting the wrong answer.

x=vot+1/2at^2
18=(1/2)(9.81)t^2
18=4.9(t)^2
sqrt(3.7)=t^2
t=1.9s
1.9s-.25s=1.65s

x=1/2(9.81)(1.7)^2
x=4.9(1.7)^2
x=14m
x=18m-14m=4m

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
azizlwl
1,065
10
It looks ok to me except for significant figures.
 
  • #3
Bahadar
4
0
Tried 4.0 tells me I got the wrong answer.
 
  • #4
Perceptron
6
0
I'll try giving this problem a shot.

We know that the initial height is 18m, and the final height for when the pelican swoops down is 0m.
From the equation Δ y= VoyT + .5gT2, we have to plug Yf as 0 m, Yi as 18m,and Voy as 0m/s.

-18m= .5gT2

√[-18/(.5g)]=T=1.9

(pelican to catch fish)-(time for fish to react)
1.9s-.25s=1.65
Yf-Yi= .5gT2

Yf-18= .5gT2
Yf=.5g(1.65)2+18=4.6m

So,it needs to see the pelican at 4.6m.

Sorry if I didn't write my English well enough.
 
Last edited:
  • #5
Bahadar
4
0
One of my previous answers I entered was 4.7m. A slight discrepancy of about .1m and it still tells me I'm wrong so not exactly sure why it's telling me that. The concept and mathematics is suppose to work.
 

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