A diver runs and dives off the end of a spring board that is 0.9 meters above the surface of the water. As she takes off from the diving board, her velocity is 5.2 m/s directed 36° above horizontal calculate:

a) the time she is in the air

s
b) the distance she lands from the board (measured horizontally from the end of the board) m
c) the maximum height she attains above the surface of the water

m

the velocity v comprises x- and y- components:

vx = 5.2 cos36° = 4.2
vy = 5.2 sin36° = 3.1

y = 0.9 + 3.1t - 4.9t^2
y=0 when t=0.85
so, she is in the air 0.85 seconds

x = 4.2t
4.2(0.85) = 3.57m
so, she lands 3.57m from the end of the board.

y is max when t = -3.1/-9.8 = 0.32
y(.32) = 1.39
max height is thus 1.39m above the water

To solve these problems, we can break down the motion of the diver into horizontal and vertical components.

Let's start by finding the time the diver is in the air.

a) Time of flight:
The vertical component of the initial velocity is given by V_y = V * sin(θ), where V is the magnitude of the velocity (5.2 m/s) and θ is the angle above the horizontal (36°).

To find the time of flight, we can use the equation:
0 = V_y * t - (1/2) * g * t^2

Where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.

Rearranging the equation, we get:
t = (2 * V_y) / g

Substituting the given values, we have:
t = (2 * 5.2 * sin(36°)) / 9.8

Using a scientific calculator, we can find the value of sin(36°) as approximately 0.5878.

t = (2 * 5.2 * 0.5878) / 9.8
t ≈ 0.626 s

Therefore, the time the diver is in the air is approximately 0.626 seconds.

b) Horizontal distance covered:
The horizontal component of the initial velocity is given by V_x = V * cos(θ), where V is the magnitude of the velocity (5.2 m/s) and θ is the angle above the horizontal (36°).

To find the horizontal distance covered, we can use the equation:
d = V_x * t

Where d is the horizontal distance covered and t is the time of flight (0.626 s in this case).

Substituting the given values, we have:
d = (5.2 * cos(36°)) * 0.626

Using a scientific calculator, we can find the value of cos(36°) as approximately 0.8090.

d = (5.2 * 0.8090) * 0.626
d ≈ 2.593 m

Therefore, the distance the diver lands from the board (measured horizontally from the end of the board) is approximately 2.593 meters.

c) Maximum height attained:
To find the maximum height attained by the diver above the surface of the water, we can use the equation:
y = y_0 + V_y * t - (1/2) * g * t^2

Where y is the vertical position, y_0 is the initial vertical position (0.9 m above the water surface), V_y is the vertical component of the initial velocity (which we already calculated as 5.2 * sin(36°)), t is the time of flight (0.626 s), and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the given values, we have:
y = 0.9 + (5.2 * sin(36°)) * 0.626 - (1/2) * 9.8 * (0.626)^2

Using a scientific calculator, we can find the value of sin(36°) as approximately 0.5878.

y = 0.9 + (5.2 * 0.5878) * 0.626 - (1/2) * 9.8 * (0.626)^2
y ≈ 2.116 m

Therefore, the maximum height attained by the diver above the surface of the water is approximately 2.116 meters.