Water from a garden hose that is pointed 21° above the horizontal lands directly on a sunbather lying on the ground 4.3 m away in the horizontal direction. If the hose is held 1.4 m above the ground, at what speed does the water leave the nozzle?

To find the speed at which the water leaves the nozzle, we need to use the principles of projectiles and the laws of motion.

First, let's break down the given information:

1. The water leaves the nozzle of the hose at an angle of 21° above the horizontal.
2. The water lands 4.3 m away in the horizontal direction.
3. The hose is held 1.4 m above the ground.

We can begin by splitting the motion into horizontal and vertical components. The horizontal motion is unaffected by gravity, so we can find the time it takes for the water to travel the horizontal distance of 4.3 m.

Horizontal motion:
1. Distance (d) = 4.3 m
2. Time (t) = ?

The horizontal distance is given by the equation: d = v * t, where v is the horizontal component of velocity. Since the horizontal velocity remains constant, we can rewrite the equation as: v = d / t.

Since the water leaves the hose at a certain angle, we must find the horizontal component of the velocity. This can be found using trigonometry. The horizontal velocity (v_h) is given by: v_h = v_initial * cos(angle), where v_initial is the initial velocity.

Now let's solve for v_initial:

1. The vertical distance (h) = 1.4 m
2. The vertical component of velocity (v_v) = ?
3. Initial velocity (v_initial) = ?

We can use the equation of motion: h = v_initial * t + (1/2) * g * t^2, where h is the vertical distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.

Since the water is going upwards first and then coming back down, the time of flight is doubled. So, we can rewrite the equation as: h = v_v * (2t) + (1/2) * g * (2t)^2.

Now, substitute the given values:

1.4 m = v_v * (2t) + (1/2) * 9.8 m/s^2 * (2t)^2

Simplifying this equation will give us a quadratic equation in terms of t:

1.4 = 2*v_v*t + 19.6*t^2

Now, we can use the fact that the water lands 4.3 m away, and both the vertical and horizontal components of motion take the same time. Therefore, h = v_h * t. Substituting the known values:

4.3 m = v_h * t

Now, we have two equations:

1.4 = 2*v_v*t + 19.6*t^2
4.3 = v_h * t

Since we have two equations and two unknowns (v_v and t), we can solve them simultaneously to find the values. There are several methods to solve these equations, including substitution, elimination, or using numerical methods like graphing.

By solving these two equations, we can find the values of v_v (vertical component of velocity) and t (time of flight). Once we have those values, we can find the initial velocity (v_initial) using the equation: v_initial = sqrt(v_h^2 + v_v^2).

Finally, we will have the initial velocity (v_initial), which is the speed at which the water leaves the nozzle.