A water slide launches a child horizontally above a swimming pool. The distance from the top of the slide to the launch point is 2.76 m. The point where the child is launched horizontally is 0.7 m above the water. How far from the launch point does the child land in the water?

You need to know two things:

1. how long does it take to fall 0.7 m
2. How fast is your horizontal velocity after falling 2.76 meters?
Then
distance = horizontal velocity * time to fall 0.7 meters = u t
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so, how long to fall 0.7 m?
0.7 = (1/2) g t^2
t = sqrt (1.4/9.81)
then how fast is the child going?
(1/2) m u^2 = m g h
u = sqrt (2 g h) = sqrt (2*9.81*2.76)

To determine how far from the launch point the child lands in the water, we can use the principles of projectile motion. We can break down the problem into two parts: vertical motion and horizontal motion.

First, let's analyze the vertical motion. We know that the child is launched horizontally, so there is no initial velocity in the vertical direction. The only force acting on the child in the vertical direction is gravity, causing the child to fall downwards.

The height from the launch point to the water surface is given as 0.7 m. Using the equation for vertical motion under constant acceleration, we can calculate the time it takes for the child to fall from this height:

h = (1/2) * g * t^2

where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Plugging in the values, we have:

0.7 = (1/2) * 9.8 * t^2

Solving for t:

t^2 = (2 * 0.7) / 9.8
t^2 = 0.14 / 9.8
t^2 = 0.0143
t ≈ √0.0143
t ≈ 0.12 s

Now, let's analyze the horizontal motion. Since the child is launched horizontally, there is no horizontal acceleration acting on the child. Thus, the horizontal velocity remains constant throughout the motion.

To determine the horizontal distance traveled, we use the equation:

d = v * t

where d is the distance, v is the horizontal velocity, and t is the time.

We need to find the horizontal velocity, which can be calculated using the equation:

v = d / t

where v is the horizontal velocity, d is the horizontal distance, and t is the time.

The horizontal distance traveled can be found by using the height from the top of the slide to the launch point, which is given as 2.76 m:

d = v * t
2.76 = v * 0.12

Solving for v:

v = 2.76 / 0.12
v ≈ 23 m/s

Now that we have the horizontal velocity, we can find the distance from the launch point to where the child lands in the water. Since the horizontal velocity is constant, the distance traveled horizontally is given by:

d = v * t

d = 23 * 0.12
d ≈ 2.76 m

Therefore, the child lands approximately 2.76 meters away from the launch point in the water.