## Physics (10th Edition)

Published by Wiley

# Chapter 5 - Dynamics of Uniform Circular Motion - Problems - Page 138: 21

#### Answer

$F_c=3482N$

#### Work Step by Step

1) The ball reached a distance $d=86.75m$. Let's call its horizontal velocity $v_x$ and the time to reach that distance $t$. We have $$v_xt=86.75 (1)$$ On the vertical, the period of time is also $t$, the ball's initial velocity $v_y$, its final velocity $-v_y$, $g=-9.8m/s^2$. We have $$-v_y=v_y+gt$$ $$2v_y=-gt=9.8t$$ $$v_y=4.9t (2)$$ Let's take velocity of release to be $v$. Since the angle of release is $41^o$, we have $v_x=v\cos41=0.75v$ and $v_y=v\sin41=0.66v$ Plug these back in (1): $vt=\frac{86.75}{0.75}=115.67$ and in (2): $v=\frac{4.9t}{0.66}=7.42t$ Solving (1) and (2), we get $v=29.3m/s$ 2) Now that we have the release speed $v$, we can find $F_c$ We also have the ball's mass $m=7.3kg$ and radius $r=1.8m$. Therefore, $$F_c=\frac{mv^2}{r}=3482N$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.