A volleyball is bumped vertically up to a height of 10 m above the player’s arms. What was the initial velocity of the ball just after leaving the player’s arms?
(1/2) m v^2 = m g h
v^2 = 2 g h
Kinetic Energy = Gravitational Potential Energy
(0.5)*mass*velocity^2 = mass*acceleration due to gravity*height above gorund
simplify by removing mass on either side => 0.5*v^2=g*h
divide by 0.5 and then square both sides to be left with => v=sqrt[2*g*h]
-196
To find the initial velocity of the ball just after leaving the player's arms, you can use the equation of motion for vertical motion:
v^2 = u^2 + 2as
Where:
- v is the final velocity (which is zero since the ball reaches its maximum height and then starts falling),
- u is the initial velocity (what we want to find),
- a is the acceleration due to gravity (-9.8 m/s^2, as the ball moves against the force of gravity), and
- s is the displacement (10 m in this case, as the ball is bumped vertically up to a height of 10 m).
First, we rearrange the equation to solve for the initial velocity (u):
u^2 = v^2 - 2as
Since the final velocity (v) is zero, the equation simplifies to:
u^2 = -2as
Now, substitute the known values:
u^2 = -2 * (-9.8 m/s^2) * 10 m
u^2 = 196 m^2/s^2
To solve for u, take the square root of both sides:
u = √(196 m^2/s^2)
u ≈ 14 m/s
Therefore, the initial velocity of the ball just after leaving the player's arms is approximately 14 m/s.