Respuesta :
Answer:
Explanation:
Impulse = change in momentum
mv - mu , v and u are final and initial velocity during impact at surface
For downward motion of baseball
v² = u² + 2gh₁
= 2 x 9.8 x 2.25
v = 6.64 m / s
It becomes initial velocity during impact .
For body going upwards
v² = u² - 2gh₂
u² = 2 x 9.8 x 1.38
u = 5.2 m / s
This becomes final velocity after impact
change in momentum
m ( final velocity - initial velocity )
.49 ( 5.2 - 6.64 )
= .7056 N.s.
Impulse by floor in upward direction
= .7056 N.s
The impulse is given by the change in the motion of the baseball, including
the change indirection.
- An expression for the impulse is I = m × (√(2·g·h₂) - (-√(2·g·h₁))
- The impulse felt by the baseball is approximately 5.81 N·s
Reason:
Known values are;
Mass of the baseball, m = 0.49 kg
Height from which the baseball is dropped, h₁ = 2.25 m
Final height to which the ball returns, h₂ = 1.38 m
(a) Required;
The expression for the impulse
Solution;
Velocity of the ball just before it reaches the floor, v₁, is given as follows;
v₁² = u² + 2·g·h₁
u = Initial velocity of the ball = 0
∴ v₁² = 2·g·h₁
v₁ = √(2·g·h₁)
The velocity with which the ball goes up, u₂, is given as follows;
v₂² = u₂² - 2·g·h₂
Where;
v₂ = The final velocity = 0
Therefore;
u₂² = 2·g·h₂
u₂ = √(2·g·h₂)
The impulse = Mass × Change in velocity
Therefore, the impulse experienced = m×(u₂ - v)
An expression for the impulse, I , is therefore;
- Impulse, I = m × (√(2·g·h₂) - (-√(2·g·h₁))
Impulse = 0.49 × (√(2 × 9.81×1.38) - (-√(2 × 9.81×2.25)) ≈ 5.81
The impulse felt by the baseball ≈ 5.81 N·s
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