Question

A water pistol aimed horizontally projects a stream of water with an initial speed of 5.80 m/s.

A) How far does the water drop in moving 1.85 m horizontally?

B) How far does it travel before dropping a vertical distance of 2.35 cm?

Answer 1

Solution :-

Given that

The horizontal speed, Vx = 5.8 m/s

1)

The horizontal distance, x = 1.85 m

We have a formula for the time as, time(t) = distance / velocity = 0.32 s

We have a formula for the displacement as

             s = Vy * t + 0.5at^2

             s = 0.5 * 9.8 * (0.32)^2 = 0.5 m

2)

Given that

     The vertical distance, y = 2.35 cm = 0.0235m

     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*0.0235/9.8) = 0.0692 sec

Then the distance is given by the formula

            s = Vx * t = 5.8 * 0.0692= 0.4 m