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 initial velocity and initial angle of the diver, we can use the equations of motion for projectile motion.
The horizontal displacement of the diver is 4 m, which means the horizontal component of the velocity remains constant throughout the motion.
The vertical motion of the diver can be analyzed using the equation:
y = y0 + v0y*t - (1/2)*g*t^2
Where:
y = vertical displacement (which is equal to the initial height of the platform, 10 m)
y0 = initial vertical position (which is 0 since the diver starts from the top of the platform)
v0y = initial vertical component of velocity (which is what we are trying to find)
t = time of flight (which is given as 2.3 s)
g = acceleration due to gravity (which is approximately 9.8 m/s^2)
Using the equation above, we can solve for v0y:
10 = 0 + v0y*2.3 - (1/2)*9.8*(2.3)^2
Simplifying the equation:
10 = 2.3*v0y - 5.3314
Rearranging the equation:
v0y = (10 + 5.3314) / 2.3
v0y = 6.776 m/s
The magnitude of the initial velocity, v0, can be found using the formula:
v0 = sqrt(v0x^2 + v0y^2)
Since the horizontal component of velocity remains constant throughout the motion, v0x is equal to the horizontal displacement of the diver, which is 4 m in this case:
v0 = sqrt((4)^2 + (6.776)^2)
v0 = 8.130 m/s
Finally, the initial angle, θ, can be found using the formula:
θ = arctan(v0y / v0x)
θ = arctan(6.776 / 4)
θ ≈ 59.04 degrees
Therefore, the diver's initial velocity is approximately 8.13 m/s, and the initial angle is approximately 59.04 degrees.