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. What is the pelican’s speed?

Speed, Vfx = 7.619 m/s

Explanation:

Vertical distance, Dx = 5.4m

Horizontal distance, Dy = 8m

Acceleration due to gravity, g = 9.8m/s²

Initial speed, Vix = 0m/s²

To find the speed, we would use the second equation of motion to find the time, t;

Dx = Vixt + ½gt²

Substituting into the equation, we have;

5.4 = 0(t) + ½(9.8)*t²

5.4 =  0 + 4.9t²

Rearranging the equation, we have;

4.9t² = 5.4

t² = 5.4/4.9

t² = 1.1020

Taking the square root of both sides;

t = 1.050 secs.

For the speed;

Dy = Vfxt

Vfx = Dy/t

Vfx = 8/1.050

Vfx = 7.619 m/s

Therefore, the speed of the pelican is 7.619 m/s

an electric circuit

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer