A diver leaves the board traveling at 3.5 m/s at an angle of 75º with respect to the horizontal. What is her speed when she hits the water, 3 meters below?
To find the diver's speed when she hits the water, we can use the concept of projectile motion. We can break down the initial velocity into its horizontal and vertical components.
Given:
Initial speed (u) = 3.5 m/s
Angle (θ) = 75º
Height (h) = 3 meters
Step 1: Find the horizontal component of velocity (u_x).
The horizontal component of velocity remains constant throughout the motion. It can be calculated using the formula:
u_x = u * cos(θ)
Substituting the given values:
u_x = 3.5 m/s * cos(75º)
Step 2: Find the vertical component of velocity at impact (v_y).
The vertical component of velocity changes due to the acceleration due to gravity. It can be calculated using the formula:
v_y^2 = u_y^2 + 2gh
Where u_y is the initial vertical component of velocity and g is the acceleration due to gravity.
Since the diver starts from rest vertically (u_y = 0), the formula simplifies to:
v_y^2 = 2gh
Substituting the given values:
v_y^2 = 2 * 9.8 m/s^2 * 3 m
v_y = sqrt(58.8) m/s
Step 3: Find the total velocity at impact (v).
The total velocity can be found using the Pythagorean theorem:
v = sqrt(v_x^2 + v_y^2)
Substituting the values:
v = sqrt(u_x^2 + v_y^2)
v = sqrt((3.5 m/s * cos(75º))^2 + (sqrt(58.8) m/s)^2)
Calculating these values will give us the speed at which the diver hits the water.