Diver jumps from a 10 m platform. Air time is 2.3 s. Horizontal displacement of the divers center of mass i 4 m. What is the divers initial velocity and initial angle?
To find the diver's initial velocity and initial angle, we can use the equations of motion.
The air time of the diver can be used to find the vertical component of the initial velocity (Vy) using the equation:
Vy = g * t
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the air time (in seconds).
In this case, Vy = 9.8 * 2.3 = 22.54 m/s (rounded to two decimal places).
The horizontal displacement of the diver's center of mass can be used to find the horizontal component of the initial velocity (Vx) using the equation:
Vx = Dx / t
where Dx is the horizontal displacement and t is the air time.
In this case, Vx = 4 / 2.3 = 1.74 m/s (rounded to two decimal places).
Now, using the components Vx and Vy, we can find the magnitude of the initial velocity (V0) using the Pythagorean theorem:
V0 = sqrt(Vx^2 + Vy^2)
In this case, V0 = sqrt(1.74^2 + 22.54^2) = 22.63 m/s (rounded to two decimal places).
Finally, to find the initial angle (θ), we can use the inverse tangent function:
θ = arctan(Vy / Vx)
In this case, θ = arctan(22.54 / 1.74) = 86.92 degrees (rounded to two decimal places).
Therefore, the diver's initial velocity is 22.63 m/s and the initial angle is 86.92 degrees.