In a friendly neighborhood squirt gun contest a participant runs at

In a friendly neighborhood squirt gun contest a participant runs at 7.8 m/s horizontally off the back deck and fires her squirt gun in the plane of her motion but 45' above horizontal. The gun can shoot water at 11 m/s relative to the barrel, and she fires the gun O.42 s 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 the horizontal and vertical components. O) At the instant she fires, the gun is 1.9 m above the level ground. How far will the water travel horizontally before landing?

3 answers

first the shooter:
horizontal velocity = us = 7.8 m/s forever
vertical velocity = vs = - g t =-9.81 t
at .42 s this is - 9.81*.42 = -4.12 m/s

now the water
horizontal velocity = u= 7.8 + 11*.707
= 15.6 m/s forever
vertical velocity =vw = -4.12 + 11*.707
= 3.66 AT .42 s
use u = 15.6 and v = 3.66 to answer part (a)
now
vw = 3.66 - 9.81(t-.42)
uw = 15.6
Hiw = initial height of water at .42 = 1.9
so create new t for time after .42 and water particle doing that:

vw = 3.66 - 9.81 t
uw = 15.6
Hiw = initial height of water at .42 = 1.9
initial x of water = .42 * 7.8=3.28
(problem does not say if interested in how far from takeoff or how far after firing)
how long in air after firing?
0 = 1.9 + 3.66 t -4.9 t^2
solve quadratic I get t = 1.1 second
so horizontal distance after discharge = 15.6 *1.1 = 17.2 meters
If you are supposed to include the horizontal 3.28 meters the gun traveledd before discharge then
17.2 +3.3 = 20.5 meters

Thank you very much Damon.

You are welcome.