It will experience centripetal accelaration.
K.E=0.5*mv²
v=square root 2ke/m
v= square root 2*8J/1 kg
v= 4 m/s
6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough
The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Given parameters
To find
Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.
In this case we have a vertical launch
y = y₀ + v₀ t - ½ g t²
Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time
With the ball in hand, its position is zero
0 = 0 + v₀ t - ½ g t²
v₀ t - ½ g t² = 0
v₀ = ½ g t
Let's calculate
v₀ = ½ 9.8 2.4
v₀ = 11.76 m / s
In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Learn more about vertical launch kinematics here:
brainly.com/question/15068914
