Consider a case in which a 60 kg diver takes off from a diving board located at a height of 10 m above the water level and enters the water at a horizontal distance of 5 m from the end of the board. If the total time that the diver remains in the air is 2.5 seconds, determine if the takeoff angle of the diver’s center of mass is greater than 45°.
To determine if the takeoff angle of the diver's center of mass is greater than 45°, we can use the laws of projectile motion.
Step 1: Calculate the vertical component of the diver's initial velocity.
Using the formula v_y = g * t (where v_y is the vertical component of the velocity, g is the acceleration due to gravity, and t is the total time in the air), we can calculate the vertical component of the initial velocity.
v_y = 9.8 m/s^2 * 2.5 s = 24.5 m/s
Step 2: Calculate the horizontal component of the diver's initial velocity.
The horizontal component of the initial velocity remains constant throughout the flight. We can calculate it using the formula S = v_x * t (where S is the horizontal distance traveled, v_x is the horizontal component of the velocity, and t is the total time in the air).
5 m = v_x * 2.5 s
v_x = 2 m/s
Step 3: Calculate the magnitude of the diver's initial velocity.
We can use the Pythagorean theorem to find the magnitude of the initial velocity (v):
v = √(v_x^2 + v_y^2)
v = √(2^2 + 24.5^2)
v ≈ 24.52 m/s
Step 4: Calculate the takeoff angle.
The takeoff angle can be determined using the formula θ = tan^(-1)(v_y/v_x).
θ = tan^(-1)(24.5/2)
θ ≈ 86.44°
Therefore, the takeoff angle of the diver's center of mass is greater than 45°.