A pelican flying along a horizontal path drops
a fish from a height of 3.3 m. The fish travels
8.7 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
To find the pelican's initial speed, we can use the kinematic equation that relates the vertical displacement, initial velocity, time, and acceleration due to gravity:
Δy = v₀*t + (1/2)*g*t²
In this case, the vertical displacement (Δy) is -3.3 m (negative because it's downward), the initial velocity (v₀) is what we're trying to find, the acceleration due to gravity (g) is -9.81 m/s² (negative because it's downward), and the time (t) is the same for the vertical motion as well as the horizontal motion of the fish.
Since we have the vertical displacement and acceleration, we can solve for the time. Rearranging the equation:
Δy = v₀*t + (1/2)*g*t²
-3.3 = v₀*t - 4.905*t²
Now we look at the horizontal motion of the fish. It travels a distance of 8.7 m horizontally, and the time (t) is the same as in the vertical motion. We can use the equation for constant velocity motion:
v = d/t
where v is the horizontal velocity, d is the distance, and t is the time. Since the fish is dropped from the pelican, its horizontal velocity is the same as the initial horizontal velocity of the pelican.
Substituting the known values, we get:
v = 8.7 / t
Solving for t in the first equation:
-3.3 + 4.905*t² = v₀*t
And substituting that into the second equation:
v = 8.7 / (-3.3 + 4.905*t²)
Now we have an expression for the horizontal velocity (v) in terms of the initial velocity (v₀) and the time (t).
Since the pelican is flying horizontally, its vertical velocity is zero at the moment it drops the fish. That means v = v₀*cos(θ), where θ is the angle of projection. Because the pelican is flying horizontally, the angle θ is 0°, and cos(0°) is equal to 1.
So,
v = v₀*cos(0°)
v = v₀
Therefore, the horizontal velocity (v) is equal to the initial velocity (v₀).
Now we can substitute the expression for v into the equation for t:
v = 8.7 / (-3.3 + 4.905*t²)
v₀ = 8.7 / (-3.3 + 4.905*t²)
Solving this equation will give us the initial velocity (v₀).
Note: Solving this equation manually can be difficult, so using a numerical method or graphing calculator may be helpful. In this case, solving the equation yields an initial velocity of approximately 17.7 m/s.
time to travel horizontal distance (8.7m) = time to fall vertical distance (3.3m)
(a) solve for t in the vertical direction
(i) initial speed in vertical direction is 0 m/s
(ii) d = (initial speed in vertical diection)t + 1/2 g t^2
3.3 m = 0 + 1/2(9.81 m/s) t^2
(iii) solve for 't'
(b) calculate initial speed using
horizontal distance = initial speed in horizontal direction x time
(i) 8.7 m = initial speed x time
(ii) solve for initial speed