A pelican flying along a horizontal path drops
a fish from a height of 4.3 m. The fish travels
8.3 m horizontally before it hits the water
below.
What was the pelican’s initial speed? The
acceleration of gravity is 9.81 m/s
2
.
Answer in units of m/s
h=gt²/2
t=sqrt(2h/g)
x=v(x) •t =>
v(x) =x/t=x/sqrt(2h/g) =
=8.3/sqrt(2•4.3/9.81)=8.9 m/s
To find the initial speed of the pelican, we can use the equations of motion. The horizontal motion of the fish does not affect the vertical motion, so we can consider the vertical motion separately.
First, let's find the time it takes for the fish to fall from a height of 4.3 m. We can use the equation of motion for free-fall:
h = (1/2) * g * t^2
where h is the height (4.3 m), g is the acceleration due to gravity (9.81 m/s^2), and t is the time. Rearranging the equation, we have:
t^2 = (2 * h) / g
t = sqrt((2 * 4.3) / 9.81)
Next, we need to find the vertical component of the initial velocity. We know that the final vertical velocity is 0 when the fish hits the water. We can use the equation of motion:
v = u + g * t
where v is the final velocity (0 m/s), u is the initial velocity in the vertical direction (which we want to find), g is the acceleration due to gravity (9.81 m/s^2), and t is the time we calculated earlier. Rearranging the equation, we have:
u = -g * t
Substituting the value of t we calculated earlier, we have:
u = -9.81 * sqrt((2 * 4.3) / 9.81)
Finally, to find the initial speed of the pelican, we need to consider the horizontal motion of the fish. The horizontal distance traveled by the fish is 8.3 m. We can use the equation of motion for constant velocity:
s = u * t
where s is the distance (8.3 m), u is the initial speed in the horizontal direction (which we want to find), and t is the time we calculated earlier. Rearranging the equation, we have:
u = s / t
Substituting the values of s and t, we have:
u = 8.3 / sqrt((2 * 4.3) / 9.81)
Calculating this expression will give us the initial speed of the pelican.