4-39 magnitude of its velocity after falling 10.0m

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Discussion Overview

The discussion revolves around a physics problem involving the motion of a water balloon thrown vertically downward from a building. Participants explore the calculations related to the balloon's speed after 2 seconds, the distance it falls in that time, and the magnitude of its velocity after falling 10.0 meters. The scope includes mathematical reasoning and application of motion formulas.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • Some participants calculate the speed after 2 seconds using the formula $v=u+at$ and arrive at a value of $27.6 \, m/s$.
  • Others provide a different calculation for the speed after 2 seconds, resulting in $25.6 \, m/s$, indicating a discrepancy in the approach taken.
  • Distance fallen in 2 seconds is calculated by some as $31.6 \, m$ using the formula $d= ut + \frac{1}{2}at^2$.
  • Participants question the magnitude of the velocity after falling 10.0 meters, with one suggesting the use of the formula $v_f^2 = v_0^2 - 2g \Delta y$ to find a value of $15.2 \, m/s$.
  • There is mention of the need to graph acceleration-time, velocity-time, and position-time relationships for the motion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations for speed and distance, as different methods yield varying results. The discussion remains unresolved regarding the correct approach to finding the magnitude of the velocity after falling 10.0 meters.

Contextual Notes

Some calculations depend on the assumptions made about the initial conditions and the interpretation of the motion equations. There are unresolved discrepancies in the application of the formulas, particularly regarding the signs and values used in the calculations.

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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