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OleMash [197]
3 years ago
8

During baseball practice, a batter hits a high flyball, and then runs in a straight line and catches it. which has the greater a

verage velocity over the time interval between hit and catch?
Physics
2 answers:
Jlenok [28]3 years ago
6 0
They have the same velocity because their displacements (shortest line from point A to point B, which is a straight line) are the same and they meet at the same time.
Trava [24]3 years ago
4 0

Answer:

The batter and the fly ball have the same average velocity, i.e. no individual velocity is greater than the other.

Explanation:

First of all, it is worthy of note that in a baseball game, when the batter hits the ball, he is expected to immediately make a run in an attempt to catch the ball, by so doing, the batter has a greater chance of catching the ball before it hits the ground  and this is exactly what the batter has done in this practice.

1. The fly ball is hit and the batter runs immediately , this means the batter and the fly ball have the same start time t.

2. The batter catches the fly ball after certain time i.e. the same end time.

3. The batter runs in a straight line to catch the ball i.e. the displacement of both batter and fly ball from hit point to catch point is the same.

MATHEMATICALLY

D_{batter}  = D_{fly ball}

t_{batter}  = t_{fly ball}

Where D = Displacement and t = time

Average velocity  V = \frac{D}{t}

Where D_{batter}  = D_{fly ball}

and

t_{batter}  = t_{fly ball}

This means V_{batter}  = V_{fly ball}

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Answer:

mass of the second ball is 0.379m

Explanation:

Given;

mass of first ball = m

let initial velocity of first ball = u₁

let final velocity of first ball  = v₁ = 0.45u₁

let the mass of the second ball = m₂

initial velocity of the second ball, u₂ = 0

let the final velocity of the second ball = v₂

Apply the principle of conservation of linear momentum;

mu₁ + m₂u₂ = mv₁ + m₂v₂

mu₁  +  0  = 0.45u₁m + m₂v₂

mu₁  = 0.45u₁m + m₂v₂ -------- equation (i)

Velocity for elastic collision in one dimension;

u₁ + v₁ = u₂ + v₂

u₁ + 0.45u₁ = 0 + v₂

1.45u₁ = v₂ (final velocity of the second ball)

Substitute in v₂ into equation (i)

mu₁  = 0.45u₁m + m₂(1.45u₁)

mu₁ = 0.45u₁m + 1.45m₂u₁

mu₁ - 0.45u₁m = 1.45m₂u₁

0.55mu₁ = 1.45m₂u₁

divide both sides by u₁

0.55m = 1.45m₂

m₂ = 0.55m / 1.45

m₂ = 0.379m

Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)

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