A diver jumps horizontally off a cliff with an initial velocity of 5.4 m/s. The diver strikes the water 4.5 s later. What is the vertical velocity of the diver upon reaching the surface of the water?
10.5m
To find the vertical velocity of the diver upon reaching the surface of the water, we need to use the equation of motion for vertical motion:
π£ = π’ + ππ‘
Where:
π£ is the final vertical velocity,
π’ is the initial vertical velocity,
π is the acceleration due to gravity (which is approximately 9.8 m/s^2),
and π‘ is the time.
In this case, the diver jumps horizontally, so there won't be any changes in the horizontal velocity. Therefore, the initial vertical velocity is 0 m/s.
Plugging the values into the equation, we have:
π£ = 0 + (9.8 m/s^2) Γ (4.5 s)
π£ = 0 + 44.1 m/s
Therefore, the vertical velocity of the diver upon reaching the surface of the water is 44.1 m/s, upward.
To find the vertical velocity of the diver upon reaching the surface of the water, we can use the equations of motion.
First, let's consider the horizontal motion of the diver. Since the diver jumps horizontally, there is no acceleration acting in the horizontal direction. Therefore, the initial horizontal velocity of 5.4 m/s remains constant throughout the motion.
Now, let's focus on the vertical motion of the diver. We can use the equation that relates displacement, initial velocity, time, and acceleration:
s = ut + (1/2)at^2
In this case, the diver starts from rest vertically, so the initial vertical velocity (u) is 0 m/s, and the acceleration due to gravity (a) is -9.8 m/s^2 (negative because it acts downward). We'll solve for the displacement (s) when the diver reaches the water.
The equation becomes:
s = (1/2)(-9.8 m/s^2)(4.5 s)^2
s = -99.225 m
So, the diver falls 99.225 meters vertically during the 4.5 second time interval.
Now, to find the vertical velocity at the surface of the water, we can use the equation:
v = u + at
In this case, the initial vertical velocity (u) is 0 m/s, the acceleration due to gravity (a) is -9.8 m/s^2, and the time (t) is 4.5 seconds.
v = 0 m/s + (-9.8 m/s^2)(4.5 s)
v = -44.1 m/s
Therefore, the vertical velocity of the diver upon reaching the surface of the water is -44.1 m/s (directed downward).