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°.
horizontal velocity=5/2.5=2m/s
hf=hi+v't-4.9t^2
0=5+v'(2.5)-4.9(2.5)^2
solve for vertical velocity v'
now, find arctan v'/2.5 is it greater orless than 45 deg.
To determine if the takeoff angle of the diver's center of mass is greater than 45°, we can use the equations of motion and kinematics.
Step 1: Calculate the initial vertical velocity component (Vy0).
The vertical velocity component at the start of the motion can be found using the equation:
Vy0 = ∆y / ∆t
Given:
Height difference (∆y) = 10 m
Total time (∆t) = 2.5 s
Vy0 = 10 m / 2.5 s
Vy0 = 4 m/s
Step 2: Calculate the horizontal velocity component (Vx) at the point of landing.
The horizontal velocity component remains constant throughout the entire motion, so it can be found using the equation:
Vx = ∆x / ∆t
Given:
Horizontal distance (∆x) = 5 m
Total time (∆t) = 2.5 s
Vx = 5 m / 2.5 s
Vx = 2 m/s
Step 3: Calculate the takeoff angle (θ) using the vertical and horizontal velocity components.
The takeoff angle can be found using the equation:
tan(θ) = Vy0 / Vx
θ = atan(Vy0 / Vx)
θ = atan(4 m/s / 2 m/s)
θ ≈ atan(2) ≈ 63.4°
Therefore, the takeoff angle of the diver's center of mass is greater than 45° (approximately 63.4°).