question_answer

A lamp is 50 ft. above the ground. A ball is dropped from the same height from a point 30 ft. away from the light pole. If ball falls a distance \[s=16{{t}^{2}}\] ft. in t seconds, then the speed of the shadow of the ball moving along the ground 1/2s later is

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)]

\[\therefore \frac{dx}{dt}=-16\frac{{{(375)}^{2}}}{1500}=-1500\,\,ft/s\]