In a friendlu neighborhood squirt gun contest a participant runs at 7.8m/s horiz

Question

In a friendlu neighborhood squirt gun contest a participant runs at 7.8m/s horizontally off the back deck and fires her squirt gun in the plane of her motion but 45 degrees above horizontal. The gun can shoot water at 11m/s relative to the barrel, an she fires the gun 0.42 seconds after leaving the deck.

A) What is the initial velocity of the water particles as seen by an observer on the ground? Give your answer in terms of horizontal and vertical components.

B)At the instant she fires, the gun is 1.9m above the level ground. How far will the water travel horizontally before landing?

Explanation / Answer

(A) V water wrt barrel = V water wrt ground - V barrel wrt ground

11 = v - 7.8

v = 18.8 m/s

(B) In vertical,

v0y = 18.8 sin45 = 13.3 m/s

ay = - 9.8 m/s^2

y = yf - yi = 0 - 1.9 = - 1.9 m

y = v0y t + ay t^2 /2

- 1.9 = 13.3 t - 4.9 t^2

t = 2.85 sec

d = (v0x) (t)

d = (18.8 cos45) (2.85)

d = 37.9 m

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