Part 1:
Janet jumps off a high diving platform with a horizontal velocity of 2.67 m/s and lands in the water 2.6 seconds later.
The acceleration is 9.8 m/s^2. How high is the platform? Answer in units of m.
Part 2:
How far from the base of the platform does she land? Answer in units of m.
Part 1:
To find the height of the platform, we can use the equation of motion:
h = v0*t + (1/2)*a*t^2
where:
h = height of the platform
v0 = initial horizontal velocity (2.67 m/s)
t = time taken to land in the water (2.6 s)
a = acceleration (9.8 m/s^2)
Substituting the given values into the equation, we get:
h = (2.67 m/s)(2.6 s) + (1/2)(9.8 m/s^2)(2.6 s)^2
Solving this equation gives us:
h = 6.942 m/s + 33.736 m
h = 40.678 m
Therefore, the height of the platform is approximately 40.678 meters.
Part 2:
To find the horizontal distance from the base of the platform where she lands, we can use the equation of motion:
d = v0*t
where:
d = horizontal distance from the base of the platform
v0 = initial horizontal velocity (2.67 m/s)
t = time taken to land in the water (2.6 s)
Substituting the given values into the equation, we get:
d = (2.67 m/s)(2.6 s)
Solving this equation gives us:
d = 6.942 m
Therefore, she lands approximately 6.942 meters from the base of the platform.
Part 1:
To find the height of the platform, we can use the kinematic equation:
h = v0y * t + (1/2) * a * t^2
Where:
h is the height of the platform,
v0y is the initial vertical velocity,
t is the time of flight, and
a is the acceleration.
In this case, we are given:
v0y = 0 m/s (the initial vertical velocity is 0 because Janet jumps horizontally)
t = 2.6 s
a = 9.8 m/s^2
So, the equation becomes:
h = 0 * 2.6 + (1/2) * 9.8 * (2.6)^2
Simplifying further:
h = (1/2) * 9.8 * 6.76
h = 49 * 6.76
h = 331.24 m
Therefore, the height of the platform is approximately 331.24 m.
Part 2:
To find the horizontal distance from the base of the platform to where Janet lands, we can use the equation:
d = v0x * t
Where:
d is the horizontal distance,
v0x is the initial horizontal velocity, and
t is the time of flight.
In this case, we are given:
v0x = 2.67 m/s (the initial horizontal velocity is given)
t = 2.6 s
So, the equation becomes:
d = 2.67 * 2.6
d = 6.942 m
Therefore, Janet lands approximately 6.942 m from the base of the platform.