A) -1500 ft/s
B) 1500 ft/s
C) -1600 ft/s
D) 1600 ft/s
Correct Answer: A
Solution :
[a] |
At time t, ball drops \[16{{t}^{2}}ft\]. distance. Therefore, |
\[y=50-16{{t}^{2}}\] ?. (1) |
Point A is the position of the falling ball at some time t. |
So, \[\frac{dy}{dt}=-32t\] |
From the figure, \[\tan \theta =\frac{y}{x}=\frac{50}{30+x}\] or |
\[y=\left( \frac{50}{30+x} \right)\cdot x\] |
\[\therefore \frac{dy}{dt}=\frac{d}{dt}\left( \frac{50x}{30+x} \right)=\frac{1500}{{{(30+x)}^{2}}}\cdot \frac{dx}{dt}\] |
or \[\frac{dx}{dt}=\frac{{{(30+x)}^{2}}}{1500}(-32t)\] |
When \[f=\frac{1}{2},y=46\] [using (1)] |
and \[x=345\] [using (2)] |
\[\therefore \frac{dx}{dt}=-16\frac{{{(375)}^{2}}}{1500}=-1500\,\,ft/s\] |
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