Janet Leapverifyar drops from a high dive platform with a horizontal velocity of 2.8 m/s. She lands in the H2O below 2.6 seconds later.
a. How high is the platform?
b. How far from the base does she land?
c. What is her final velocity in the vertical direction?
a. h=g*t
h=9.8*2.6 meters
b. distance=horizveloc*time=2.8*2.6 m
c. final Ke= iniitla KE+initial PE
1/2 m vf^2=1/2 m (2.8)^2 + m g habove
To solve the given problem, we can use the equations of motion under constant acceleration. Let's break down each part of the problem and find the answers step by step:
a. To determine the height of the platform from which Janet Leapverifyar jumps, we can use the equation of motion for vertical displacement:
Δy = v₁y * t + (1/2) * a * t²
Where:
Δy = Vertical displacement (height of the platform)
v₁y = Initial vertical velocity (0 m/s, as Janet jumps straight down)
t = Time taken to reach the water (2.6 seconds, given in the question)
a = Acceleration due to gravity (-9.8 m/s², assuming downward as the positive direction)
Substituting the values into the equation, we get:
Δy = 0 * 2.6 + (1/2) * (-9.8) * (2.6)²
Simplifying, we find:
Δy = -33.572 m
Since the height cannot be negative, we take the absolute value of Δy to get:
Δy = 33.572 m
Therefore, the height of the platform is approximately 33.572 meters.
b. To determine the horizontal distance from the base where Janet lands, we can use the equation of motion for horizontal displacement:
Δx = v₁x * t
Where:
Δx = Horizontal displacement (distance from the base)
v₁x = Initial horizontal velocity (2.8 m/s, given in the question)
t = Time taken to reach the water (2.6 seconds, given in the question)
Substituting the values into the equation, we get:
Δx = 2.8 * 2.6
Simplifying, we find:
Δx = 7.28 m
Therefore, Janet lands approximately 7.28 meters from the base.
c. Janet's final velocity in the vertical direction can be determined using the equation of motion:
v₂y = v₁y + a * t
Where:
v₂y = Final vertical velocity
v₁y = Initial vertical velocity (0 m/s, as Janet jumps straight down)
a = Acceleration due to gravity (-9.8 m/s², assuming downward as the positive direction)
t = Time taken to reach the water (2.6 seconds, given in the question)
Substituting the values into the equation, we get:
v₂y = 0 + (-9.8) * 2.6
Simplifying, we find:
v₂y = -25.48 m/s
Therefore, Janet's final velocity in the vertical direction is approximately -25.48 m/s. The negative sign indicates that her final velocity is directed downward.