m = mass of the ball being raised
h = height to which the ball is raised = 0.5 m
v₀ = initial speed given to the ball = 3 [tex]\frac{m }{s}[/tex]
v = final speed of the ball at its lowest position = ?
using conservation of energy
final kinetic energy at the lowest point = initial kinetic energy + initial potential energy
(0.5) m v² = (0.5) m v₀² + mgh
dividing each term by "m"
(0.5) v² = (0.5) v₀² + gh
inserting the values
(0.5) v² = (0.5) (3)² + (9.8) (0.5)
v = 4.34 m/s
While undergoing the simple harmonic motion of pendulum, the speed of ball at lowest position is 4.33 m/s.
Given data:
Height raised by ball is, [tex]h=0.5 \;\rm m[/tex].
Initial speed of ball is, [tex]u=3.0 \;\rm m/s[/tex].
Let m be the mass of pendulum bob (ball). Then, apply the conservation of energy as,
Kinetic energy at lowest point = Kinetic energy at highest point + Potential energy at higher point
[tex]K = K' +P'\\\dfrac{1}{2}mv^{2}= \dfrac{1}{2}mu^{2} +mgh[/tex]
Here, g is the gravitational acceleration and v is the speed of ball at lowest point.
Solving as,
[tex]\dfrac{1}{2}v^{2}= \dfrac{1}{2}u^{2} +gh\\\dfrac{1}{2}v^{2}= \dfrac{1}{2} \times 3^{2} +(9.8 \times 0.5)\\ v = 4.33 \;\rm m/s[/tex]
Thus, we can conclude that the speed of the ball at its lowest position is 4.33 m/s.
Learn more about conservation of energy here:
https://brainly.com/question/13010190?referrer=searchResults
