MHB 4-39 magnitude of its velocity after falling 10.0m

AI Thread Summary
The discussion focuses on the calculations related to a water balloon thrown downward from a building. After 2 seconds, the speed of the balloon is calculated to be 27.6 m/s, and it falls a distance of 31.6 meters in that time. The magnitude of its velocity after falling 10.0 meters is determined to be 15.2 m/s. The calculations utilize motion formulas for speed and distance, considering the initial velocity and acceleration due to gravity. The discussion also highlights the importance of correctly applying these formulas for accurate results.
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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