Verified answer

Let the nozzle speed = v(i)

vertical component of nozzle speed = v(i) sin 20°

horizontal component of nozzle speed = v(i) cos 20° ... [and ignoring air resistance that stays constant]

The nozzle is 1.4 m above the ground ... so the water ends up 1.4 m below where it started when it hits the sunbather

so s = -1.4 m ... [taking UP as the positive direction]

a = -g = -9.8 m/s²

horizontal range to sunbather = 4.6 m

have to find the time it takes the water to hit the sunbather:

Consider the vertical motion

s = v(i)t + (1/2)at²

-1.4 = v(i) sin 20° t + (1/2) * (-9.8) t²

-1.4 = v(i) sin 20° t - 4.9t² ... [eqn 1]

Now consider the horizontal motion:

Horizontal range = horizontal velocity x time of flight

4.6 = v(i) cos 20° t

t = 4.6 / [v(i) cos 20°] ... [eqn 2]

subs [eqn 2] into [eqn 1] → -1.4 = v(i) sin 20° * (4.6 / [v(i) cos 20°]) - 4.9 * (4.6 / [v(i) cos 20°])²

-1.4 = [4.6 v(i) sin 20° / v(i) cos 20°] - (4.9 * 4.6²) / [v(i) cos 20°]²

-1.4 = 4.6 tan 20° - 103.684 / [v(i)² cos² 20°]

103.684 / [v(i)² cos² 20°] = 4.6 tan 20° + 1.4

103.684 = [4.6 tan 20° + 1.4] * [v(i)² cos² 20°]

v(i)² = 103.684 / {[4.6 tan 20° + 1.4] * cos² 20°]}

v(i) = √[103.684 / {[4.6 tan 20° + 1.4] * cos² 20°]}]

v(i) = 6.2 m/s (to 2 sig figs)

btw ... I've left most of the actual calculation to the end to reduce the effect of rounding errors