If the height of the water slide in the is = 3.9m , and the person's initial speed at point A is 0.60 m/s , what is the horizontal distance between the base of the slide and the splashdown point of the person?

also sorry to add, the drop off point from slide is 1.5m from ground.

To find the horizontal distance between the base of the slide and the splashdown point of the person, we need to consider the laws of projectile motion.

First, we'll break down the initial velocity of the person into its vertical and horizontal components. The vertical component of the velocity does not affect the horizontal distance traveled, so we'll focus on the horizontal component.

Given that the person's initial speed at point A is 0.60 m/s, this represents the horizontal component of the velocity. Therefore, the initial horizontal velocity, Vx, is equal to 0.60 m/s.

Next, we use the equation of motion: d = Vx * t, where d is the horizontal distance, Vx is the horizontal velocity, and t is the time taken.

To find the time taken, we need to consider the vertical motion of the person. Since the water slide is inclined, the person will experience both vertical and horizontal motion.

Using the equation of motion for vertical motion: h = Vyi * t + (1/2) * g * t^2, where h is the height of the water slide, Vyi is the initial vertical velocity (which is 0 since the person starts from rest vertically), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.

Rearranging the equation, we have: t = sqrt((2 * h) / g)

Plugging in the values: t = sqrt((2 * 3.9) / 9.8) ≈ 0.89 seconds (rounded to two decimal places).

Finally, we can calculate the horizontal distance using d = Vx * t:
d = 0.60 m/s * 0.89 s ≈ 0.53 meters (rounded to two decimal places).

Therefore, the horizontal distance between the base of the slide and the splashdown point of the person is approximately 0.53 meters.