A white-crowned sparrow flying hoizontaly with a speed of 1.80 m/s folds its win

Question

A white-crowned sparrow flying hoizontaly with a speed of 1.80 m/s folds its wings and begins to drop in freefall.
a. How far does the sparrow fall after traveling a horizontal distance of 0.500 m?
b. If the sparrow's initial speed is increased does the distance of fall increase, decrease, or stay the same?

Explanation / Answer

(a) The time taken to travel a horizontal distance is X = vx * t ==> t = X / vx = 0.5 / 1.80 = 0.2777 s The vertical fall height of the bird will drop to y = Vyt + 0.5 a t^2 y = 0 + 0.5* -9.82* ( 0.2777^2) = - 0 .326m (b) The distance will decrease because the amount of time it takes for the bird to fly the 0 .5m will decrease resulting in a decrease in the amount the bird drops.

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