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3 years ago

The boy on the tower throws a ball 20m downrange. What is his pitching speed?

Physics

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

The pitching speed of the ball is 19.7 m/s

Explanation:

Here, we can use the third equation of motion, $v^{2} = u^{2} - 2as$

whereas v represents the final velocity, u represents initial velocity, a is the acceleration due to gravity and s is the displacement or distance an object traveled

Here, the initial velocity of the the ball is given as zero and the acceleration due to gravity is 9.8 , the distance 's' is given as 20 m

Using the equation, $v^{2} = 2 * 9.8 * 20 = 392\\v = \sqrt{392} = 19.7m/s$

Hence, the pitching speed of the ball is 19.7 m/s

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Answer:

5.5 x 10^5 N/C

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3 years ago

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

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Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

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3 years ago

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3 years ago