If a high-diver performs a handstand dive (falls vertically) from a platform and enters the water 2.04 seconds later, through what displacement must the diver have fallen? Express your answer from the platform to the ground and indicate the displacment in this direction. Tip: Remember that any motion downwards is negative (-ve). Note 1: The units are not required to be included in this instance.
Well, that's quite a tricky question. So, let's dive into it, shall we?
If the diver falls vertically during a handstand dive, we can assume that their initial velocity is zero. We can use the equation of motion:
displacement = (initial velocity × time) + (0.5 × acceleration × time^2)
Since the diver is falling downwards, we can take acceleration due to gravity (g) as -9.8 m/s^2.
So, putting the values in, we have:
displacement = (0 × 2.04) + (0.5 × (-9.8) × (2.04)^2)
Now, let's calculate this without clowning around too much.
displacement = -20.03
Therefore, the displacement is -20.03 units (indicating downward direction) from the platform to the ground. And remember, gravity likes to pull us down, so it's a good idea not to keep too many things up in the air!
To find the displacement of the diver, we can use the formula for displacement:
Displacement = Initial velocity × time + (1/2) × acceleration × time^2
In this case, the diver starts from rest (initial velocity = 0) and falls vertically, so the acceleration will be equal to the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it is downwards). Plugging in the given values:
Displacement = 0 × 2.04 + (1/2) × (-9.8) × (2.04)^2
Simplifying the equation:
Displacement = -4.9 × (2.04)^2
Calculating:
Displacement = -4.9 × 4.1616
Displacement ≈ -20.3544
Therefore, the displacement of the diver is approximately -20.3544 units (downwards).
To calculate the displacement of the high-diver, we can use the equation of motion:
d = v0t + (1/2)gt^2
where:
d is the displacement (what we need to find)
v0 is the initial velocity (which is 0 since the diver starts from rest)
t is the time taken (given as 2.04 seconds)
g is the acceleration due to gravity (-9.8 m/s^2, as downward motion is negative)
Plugging in the given values, we have:
d = 0(2.04) + (1/2)(-9.8)(2.04)^2
Simplifying the equation:
d = 0 + (-4.9)(2.04)^2
d = -4.9(4.1616)
d = -20.36424
Therefore, the high-diver must have fallen a displacement of -20.36424 units (in the downward direction) from the platform to the ground.
a =-g = -9.81 m/s^2
v = Vi + a t = 0 - 9.81 t
h = Hi + Vi t + (1/2) a t^2 = Hi + 0 -4.9 t^2
h = 0 at water
Hi = 4.9 t^2
Hi = 4.9 (2.04)^2 = 20.4 meters
d = displacement from top = -20.4 m