Question

A pelican flying along a horizontal path drops a fish from a height of 12.3 m. the fish travels 16.0m horizontally before it hits the water below. What is the Pelican's initial speed?

Answer 1

10.1 m/s

Step-by-step explanation:

Let the coordinates of the pelican's position be (x, y).

Now, since it is dropped from a height of 12.3 m, then initial velocity is zero and using Newton's 2nd law of motion, we have;

12.3 = 0 - ½gt²

½gt² = 12.3

t² = 12.3 × 2/9.8

t² = 2.51

t = √2.51

t = 1.5843 s

Now, it travels 16.0m horizontally before it hits the water below. Using the formula, distance = speed × time, we have;

16 = v_x•t

Thus;

16 = v_x × 1.5843

v_x = 16/1.5843

v_x = 10.1 m/s

Now, the velocity in the vertical y-direction is zero.

Thus;

v = √((v_x)² + (v_y)²)

v = √((10.1² + 0²))

v = 10.1 m/s