<p>&#160;<span style=\"color: red;\">0.13</span>&#160;kg baseball moving at&#160

Question

<p>&#160;<span>0.13</span>&#160;kg baseball moving at&#160;<span>+</span><span>20.80</span>&#160;m/s is slowed to a stop by a catcher who exerts a constant force of&#160;<span>-</span><span>377</span>&#160;N.&#160;</p>
<div class="indent">How long does it take this force to stop the ball?<br /><span class="qTextField"><input id="RN_555453_12_0_555993" dir="ltr" name="RN_555453_12_0_555993" size="10" type="text" /></span>&#160;s<br />How far does the ball travel before stopping?<br /><span class="qTextField"><input id="RN_555453_12_1_555993" dir="ltr" name="RN_555453_12_1_555993" size="10" type="text" /></span>&#160;m</div>

Explanation / Answer

Force exerted = -377 N

So deceleration provided by the catcher = F / m = - 377 / 0.13 = - 2900 m/s2

Using 1st equation of motion, v = u + at

v= 0, u = 20.80 m/s

So t = -20.80 / -2900 = 7.17 x 10-3 sec

Distance travelled by ball can be found out using v2- u2 = 2aS

So 02 - 20.82 = 2 x -2900 x S

S = 0.0745 m = 7.45 cm

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