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Liono4ka [1.6K]

3 years ago

15

Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea

chball's velocity at the moment it was tossed into the air?

2 answers:

notka56 [123]3 years ago

3 0

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

$v=u+at$

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

$0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s$

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

charle [14.2K]3 years ago

3 0

Answer:4.9

Explanation:

Khan academy

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kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

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The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

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