Question

A water pistol aimed horizontally projects a stream of water with an initial speed of 6.30 m/s. (a) How far does the water drop in moving 1.15 m horizontally? (b) How far does it travel before dropping a vertical distance of 2.20 cm?

Answer 1

The horizontal speed, Vx = 6.30 m/s

1)

The horizontal distance, x = 1.15 m

We have a formula for the time as, time(t) = distance / velocity = 1.15 /6.3 = 0.1825 s

We have a formula for the displacement as

             s = Vy * t + 0.5at^2

             s = 0.5 * 9.8 * (0.1825)^2 = 0.16327 m

2)

Given that

     The vertical distance, y = 2.20 cm = 0.022 m

     To find out the time

       y = Vy * t + 0.5at^2

From the above we have

        y = 0.5at^2

        t = sqrt (2y/a) = sqrt(2*0.022/9.8) = 0.0670059394 sec

Then the distance is given by the formula

            s = Vx * t = 6.3 * 0.0670059394 = 0.422137418 m