Applications Of Derivatives Question 26
Question: 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=16t^{2} $ ft. in t seconds, then the speed of the shadow of the ball moving along the ground 1/2s later is
Options:
A) -1500 ft/s
B) 1500 ft/s
C) -1600 ft/s
D) 1600 ft/s
Show Answer
Answer:
Correct Answer: A
Solution:
[a] At time t, ball drops $ 16t^{2} $ feet. distance. Therefore, $ y=50-16t^{2} $ …………. (1) Point A is the position of the falling ball at some time t. So, $ \frac{dy}{dt}=-32 $ From the figure, $ \tan \theta =\frac{y}{x}=\frac{50}{30+x} $ or $ y=( \frac{50}{30+x} )\cdot x $
$ \therefore \frac{dy}{dt}=\frac{d}{dt}( \frac{50x}{30+x} )=\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 $
BETA
