A swimmer on a water-slide ,starts from rest at the top of the slide and leaves the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.500 s after leaving the slide. Ignoring friction and air resistance, find the height of the start point of the slide above the water.
the horizontal speed is 5m/0.5s = 10 m/s
Now just use the fact that PE lost = KE gained, so
gh = 1/2 v^2
plug in your numbers and solve for h. That is the height of the slide above where the person shot out.
So now add to that the distance the person dropped in 0.5s (s = 1/2 gt^2)
To find the height of the start point of the slide above the water, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
s = displacement (horizontal distance traveled after leaving the slide) = 5.00 m
u = initial velocity (horizontal velocity at the end of the slide)
t = time taken after leaving the slide = 0.500 s
a = acceleration (horizontal acceleration)
Since there is no horizontal acceleration and the swimmer leaves the slide traveling horizontally, the initial velocity (u) will remain constant.
Therefore, the equation becomes:
s = ut
Rearranging the equation to solve for u:
u = s / t
Plugging in the values:
u = 5.00 m / 0.500 s
u = 10.0 m/s
Now, we can use the vertical motion equations to find the height of the start point of the slide:
v = u + at
Since the swimmer starts from rest at the top of the slide, the initial vertical velocity (v) will be zero. Also, the only force acting on the swimmer in the vertical direction is due to gravity, which provides a constant downward acceleration (a).
Therefore, the equation becomes:
0 = 0 + at
Simplifying:
at = 0
Rearranging the equation to solve for t:
t = 0 / a
t = 0 s (since the swimmer hits the water 0.500 s after leaving the slide)
Now, we can use the vertical displacement equation:
s = ut + (1/2)at^2
Since the initial vertical velocity (u) is zero and the time (t) is zero, the equation simplifies to:
s = (1/2)at^2
The displacement (s) in the vertical direction is equal to the height (h) of the start point of the slide above the water.
Therefore, the equation becomes:
h = (1/2)at^2
Plugging in the value of the acceleration due to gravity (a = 9.8 m/s^2) and the time (t = 0.500 s):
h = (1/2)(9.8 m/s^2)(0.500 s)^2
h = (1/2)(9.8 m/s^2)(0.250 s^2)
h = 1.225 m
Therefore, the height of the start point of the slide above the water is 1.225 meters.
To find the height of the start point of the slide above the water, we can use the kinematic equation for horizontal motion:
Δd = v₀ * t
Where:
Δd is the horizontal displacement (distance from the end of the slide to where the swimmer hits the water).
v₀ is the initial horizontal velocity (the speed of the swimmer when leaving the slide).
t is the time taken for the swimmer to travel the horizontal displacement.
From the information given in the question, we have:
Δd = 5.00 m
t = 0.500 s
We need to find v₀, the initial horizontal velocity.
Since the swimmer starts from rest at the top of the slide, the initial vertical velocity (v₀_y) is 0 m/s.
Using the vertical motion equation:
h = v₀_y * t + (1/2) * g * t²
Where:
h is the height of the start point of the slide above the water.
g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the v₀_y is 0 m/s, the equation simplifies to:
h = (1/2) * g * t²
Plugging in the values:
h = (1/2) * 9.8 m/s² * (0.500 s)²
h ≈ 1.225 m
Therefore, the height of the start point of the slide above the water is approximately 1.225 meters.