4-39 magnitude of its velocity after falling 10.0m

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SUMMARY

The discussion focuses on the physics of a water balloon thrown vertically downward with an initial speed of 6.00 m/s. Using the motion formula \(v = u + at\), the speed after 2.00 seconds is calculated to be 27.6 m/s. The distance fallen in the same time frame is determined using \(d = ut + \frac{1}{2}at^2\), yielding a total fall of 31.6 m. Finally, the magnitude of the velocity after falling 10.0 m is computed using \(v_f^2 = v_0^2 - 2g\Delta y\), resulting in a final speed of 15.2 m/s.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of motion under gravity, particularly in free fall scenarios.

karush
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a) what is its speed after falling to 2.00s
motion formula $v=u+at$
so
$v = 6.00 \dfrac{m}{s}
+ 9.81 \dfrac{m}{s^{\cancel{2}}}{2.00 \cancel{s}}
= 27.6 \dfrac{m}{s}$
b) How far does it fall in 2.00s
distance formula $d= ut + \dfrac{1}{2}at^2$
so
$d=6.00\dfrac{m}{\cancel{s}}\cdot 2.00\cancel{s}
+ \dfrac{1}{2}\cdot 9.81 \dfrac{m}{s^{\cancel{2}}}{(2.00s)^2}
=12.00m+19.62m=31.6m$
c) What is the magnitude of its velocity after falling 10.0m?
graph $a-t, v-t, \textit{ and } y-t$ graphs for the motion.

ok don't have book answer to these and a little ?? on c} and graph
 
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karush said:
a) what is its speed after falling to 2.00s
motion formula $v=u+at$
so
$v = 6.00 \dfrac{m}{s}
+ 9.81 \dfrac{m}{s^{\cancel{2}}}{2.00 \cancel{s}}
= 27.6 \dfrac{m}{s}$
b) How far does it fall in 2.00s
distance formula $d= ut + \dfrac{1}{2}at^2$
so
$d=6.00\dfrac{m}{\cancel{s}}\cdot 2.00\cancel{s}
+ \dfrac{1}{2}\cdot 9.81 \dfrac{m}{s^{\cancel{2}}}{(2.00s)^2}
=12.00m+19.62m=31.6m$
c) What is the magnitude of its velocity after falling 10.0m?
graph $a-t, v-t, \textit{ and } y-t$ graphs for the motion.

ok don't have book answer to these and a little ?? on c} and graph

What information does the original problem statement give prior to asking parts (a), (b), and (c)?
 
4-39 A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's had with a speed of $6.00 m/s$ Air resistance may by ignored, so the water balloon is in free fall after it leaves the throwers hand

a) what is its speed after falling to 2.00s
motion formula is $v=u+at$
so
$v = 6.00 \dfrac{m}{s}
+ 9.81 \dfrac{m}{s^{\cancel{2}}}{2.00 \cancel{s}}
= 27.6 \dfrac{m}{s}$
b) How far does it fall in 2.00s
distance formula $d= ut + \dfrac{1}{2}at^2$
so
$d=6.00\dfrac{m}{\cancel{s}}\cdot 2.00\cancel{s}
+ \dfrac{1}{2}\cdot 9.81 \dfrac{m}{s^{\cancel{2}}}{(2.00s)^2}
=12.00m+19.62m=31.6m$
c) What is the magnitude of its velocity after falling 10.0m?
graph $a-t, v-t, \textit{ and } y-t$ graphs for the motion.

c) ?
 
4-39 A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's had with a speed of [FONT=MathJax_Main]6.00[FONT=MathJax_Math]m[FONT=MathJax_Main]/[FONT=MathJax_Math]s Air resistance may by ignored, so the water balloon is in free fall after it leaves the throwers hand

(a) $v(t) = v_0 - gt$

$v(2) = -6 - g(2) = -25.6 \, m/s$

speed is $|v| = |-25.6| = 25.6 \, m/s$

(b) $\Delta y = v_0 t - \dfrac{1}{2}gt^2$

$\Delta y = -6(2) - \dfrac{1}{2}g(2^2) = -31.6 \, m$

(c) $v_f^2 = v_0^2 - 2g \Delta y$

$v_f = -\sqrt{(-6)^2 - 2g(-10)} = -15.2 \, m/s$

$|v_f| = 15.2 \, m/s$
 
ok, much mahalo

look like I didn't subtract :eek:
 

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