A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. The acceleration of gravity is 9.81 m/s2.

a) What was the pelican's initial speed?

b) If the pelican was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?

a) V = 6.25 m/s

b) d = 4.6 m

Explanation:

a) The pelican's initial speed is equal to the horizontal speed of the fish. So we can calculate Vx of the fish knowing the fact that it travelled 8.0 m, using the formula for an Uniformly Accelerated Motion (because the fish is freefalling) to calculate the time the fish was falling:

D (t) = 0.5*a*t² + V₀*t + e₀

In this case, V₀ and e₀ are zero, a is gravity's acceleration and D (t) is 8.0 m

8.0 m = 0.5 * 9.81m/s² * t²

t² = 1.63 s²

t = 1.28 s

Thus Vx of the fish is:

8.0 m / 1.28 s = 6.25 m/s

And that's the same initial speed of the pelican.

b) The pelican is traveling at the same speed, so Vx of the fish remains the same, 6.25 m/s. First we calculate the time again:

D (t) = 0.5*a*t²

2.7 m = 0.5 * 9.81m/s² * t²

t² = 0.55 s²

t = 0.74 s

Now we use the formula V = d/t and solve for t:

6.25 m/s = d / 0.74 s

d = 4.6m