A white-crowned sparrow flying horizontally with a speed of 1.65 m/s folds its wings and begins to drop in free fall.

(a) How far does the sparrow fall after traveling a horizontal distance of 0.550 m?
(b) If the sparrow's initial speed is increased, does the distance of fall increase, decrease, or stay the same?

time to cover .550m = .550m / 1.65m/s = 1/3 s

distance fallen in 1/3 s = 4.9 * (1/3)^2 = 0.54m

since s(t) = Vo*t + 1/2 at^2, if its initial speed changes, its distance does so in like manner. Think about it. If you throw a rock down, it will fall farther than if you just drop it.

(a) Well, I guess we could say the sparrow is really falling for the "feather droppin'" technique. To find out how far our little aerial acrobat is going to drop, let's assume gravity has its usual 9.8 m/s² act going on.

Now, since the sparrow is in free fall, the horizontal motion won't affect its vertical motion. That means we can treat the vertical drop as if it were just a normal projectile motion. The time it takes for the sparrow to move 0.550 m horizontally is simply the distance divided by its horizontal speed, which is 1.65 m/s in this case.

So, time taken = 0.550 m / 1.65 m/s = 0.333 seconds (approximately).

Since we're dealing with a vertical free fall, we can use the equation d = 1/2 * g * t² to find out how far the sparrow falls. Plugging in the value of time, we get:

Vertical distance = 1/2 * 9.8 m/s² * (0.333 s)² = 0.549 m (approximately).

So, the sparrow falls approximately 0.549 meters after traveling a horizontal distance of 0.550 m. Quite the gravity-defying drop!

(b) Now, if we increase the sparrow's initial speed before it folds its wings and starts its free fall, the distance of fall would actually increase. You see, the more initial speed it has, the greater the horizontal distance it would cover before it starts falling freely. And since the horizontal motion doesn't affect the free fall part, the distance of fall would be independent of the horizontal distance covered. So, crank up that initial speed, and watch the sparrow take the plunge over a longer distance! Just don't forget to give it a parachute. Safety first!

To solve this problem, we need to consider the horizontal and vertical motions of the sparrow independently.

(a) To find the distance the sparrow falls after traveling a horizontal distance of 0.550 m, we can use the equations of motion in free fall.

The sparrow is initially flying horizontally with a constant velocity, so there is no horizontal acceleration. Therefore, the horizontal distance it travels does not affect the vertical motion.

In the vertical direction, the sparrow is in free fall. Assuming there is no air resistance, the only force acting on the sparrow in the vertical direction is gravity. The acceleration due to gravity is approximately 9.8 m/s^2, directed downwards.

Since the sparrow is in free fall, we can use the kinematic equation:

Δy = v₀y * t + 0.5 * a * t^2

where Δy is the vertical distance the sparrow falls, v₀y is the initial vertical velocity (0 m/s since the sparrow folds its wings and starts to drop), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

To find the time, we can use the horizontal distance and horizontal velocity:

t = Δx / v₀x

where Δx is the horizontal distance the sparrow travels (0.550 m) and v₀x is the initial horizontal velocity (1.65 m/s).

Plugging in the values, we get:

t = 0.550 m / 1.65 m/s = 0.333 s

Now we can substitute the time into the first equation to find the vertical distance:

Δy = 0 * 0.333 s + 0.5 * (-9.8 m/s^2) * (0.333 s)^2
= -0.544 m

Therefore, the sparrow falls a vertical distance of approximately 0.544 meters.

(b) If the sparrow's initial speed is increased, it will not change the distance of the fall. The acceleration due to gravity remains the same regardless of the initial speed. Therefore, the sparrow will still fall the same vertical distance for a given horizontal distance, assuming no other factors like air resistance come into play.

Oops. Since the sparrow's speed is given as horizontal, it will fall the same vertical distance no matter how fast it is going horizontally.