What Minimum Height Fish Must Spot Pelican to Escape?

[stacker]

Homework Statement

If it takes a fish 0.19s to perform evasive action, at what minimum height must it spot the pelican to escape?

[Suppose a pelican starts its dive from a height of 15.9 m and cannot change its path once committed]

If it takes a fish 0.19s 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. (Express your answer using two significant figures.

None of the following answers are correct:
13
12
14
19
5.3
5.8
20
1.95
4.9

Homework Equations

X = X (initial) + V (initial) * T + 1/2*A*T^2, which is simplified to X (initial) = 1/2*A*T^2

X (initial) = 1/2*A*T^2 = 1/2*9.8*1.9^2 = 19.6

X = X (initial) + [V^2-V(initial)^2]/2A

The Attempt at a Solution

I'v used the equation X = X (initial) + V (initial) * T + 1/2*A*T^2, which is simplified to X (initial) = 1/2*A*T^2.
-15.9 meters = 1/2*(-9.8)*T^2 and solved for T;
T = 1.80 seconds
Total time needed for evasion is 1.80 s + 0.19 s = 1.99 s or 2.0 seconds and I used 1.80 s - 0.19 s = 1.61 s.
I plugged in 2.0 seconds into "X (initial) = 1/2*A*T^2 = 1/2*9.8*1.9^2 = 19.6" and I tried "= 1/2*9.8*1.61^2 = 12.7 or 13."

Both were found to be wrong.

If you were to determine velocity of pelican, you would use " X = X (initial) + [V^2-V(initial)^2]/2A".
-15.9 meters * 2 (-9.8)= V^2 ; V = square root of (311.64) = 17.65 m/s or ~17.7 m/s.

But I think velocity is irrelevant.

 

[Kurdt]

Homework Statement

If it takes a fish 0.19s to perform evasive action, at what minimum height must it spot the pelican to escape?

[Suppose a pelican starts its dive from a height of 15.9 m and cannot change its path once committed]

If it takes a fish 0.19s 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. (Express your answer using two significant figures.

None of the following answers are correct:
13
12
14
19
5.3
5.8
20
1.95
4.9

Homework Equations

X = X (initial) + V (initial) * T + 1/2*A*T^2, which is simplified to X (initial) = 1/2*A*T^2

X (initial) = 1/2*A*T^2 = 1/2*9.8*1.9^2 = 19.6

X = X (initial) + [V^2-V(initial)^2]/2A

The Attempt at a Solution

I'v used the equation X = X (initial) + V (initial) * T + 1/2*A*T^2, which is simplified to X (initial) = 1/2*A*T^2.
-15.9 meters = 1/2*(-9.8)*T^2 and solved for T;
T = 1.80 seconds
Total time needed for evasion is 1.80 s + 0.19 s = 1.99 s or 2.0 seconds and I used 1.80 s - 0.19 s = 1.61 s.
I plugged in 2.0 seconds into "X (initial) = 1/2*A*T^2 = 1/2*9.8*1.9^2 = 19.6" and I tried "= 1/2*9.8*1.61^2 = 12.7 or 13."

Both were found to be wrong.

If you were to determine velocity of pelican, you would use " X = X (initial) + [V^2-V(initial)^2]/2A".
-15.9 meters * 2 (-9.8)= V^2 ; V = square root of (311.64) = 17.65 m/s or ~17.7 m/s.

But I think velocity is irrelevant.

All you're looking for really is the distance the pelican travels in the final 0.19 seconds of its dive.

 

[quietrain]

let s be the distance pellican travel in the last 0.19s of its dive ( i assume its free fall, so a = 9.8)

so

s = ut + (1/2) at²
s = u(0.19) + (1/2) (9.8) (0.19²) =====> equation 1

now to find u, which is the speed of pellican at the start of the last 0.19s of its dive,

we use

v'² = u'² + 2as', =====> equation 2

where now, v' is your u,
u' is initial speed = 0 , assume pellican dives from rest
s' is distance covered from start till the last 0.19s of the dive, = (15.9 - s)
a =9.8

so substituting equation 2 into 1,

s = [2a(15.9-s)]^1/2 (0.19) + (1/2) (9.8) (0.19²)

solving for s, i think you will get around 3m (you should get a more accurate number to 2s.f as specified by question)