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How tall is a tower if it casts a shadow 65 ft. long at the same time that a building 160 ft. tall casts a shadow 130 ft. long? I tried using the hypotenuse formula but didn't get the right answer. Help.

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2  
You need to use similar triangles. –  Arturo Magidin Jun 15 '12 at 5:18
    
$\frac{160}{130} = \frac{x}{65}$ so $x = 80$. "You need to use similar triangles." –  William Jun 15 '12 at 5:23
    
Thank you very much, I was using the wrong relationships for the proportion. –  Jorge Jun 15 '12 at 5:30
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Of course, the canonical way of finding the height of a building is to offer a barometer to the building manager. –  Arturo Magidin Jun 15 '12 at 5:36

3 Answers 3

As Arturo said, you need to use similar triangles. Here’s a picture to get you started: enter image description here

The building and its shadow, together with the hypotenuse joining their ends, form a triangle similar to the one formed by the tower and its shadow. You know the lengths of both shadows and the height of the tower; now just use the fact that the large triangle is just a scaled up version of the smaller one: every side has been multiplied by the same amount.

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If $\alpha$ is the angle that the Sun rays make with the Earth, then $\tan \alpha = \frac{h}{s} = \frac{h_1}{s_1}$, where $h,s$ are the height and shadow lengths of the first building and index one stands for the second building. Then, $h = \frac{s}{s_1} h_1 =\frac{65}{130} 160 = 80 ft$.

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Toss a firecracker on to the street from the roof. It will fall at 30 feet per second, the speed of gravity. When it hits the ground the startled pedestrians below will likely swear in surprise. Their swear words will float back up to you at 1100 feet per second, the speed of sound. Therefore the length of time for you to hear swearing after you chuck the firecracker will be:

height/30 seconds + height/1100

=

1100 x height / 33000 + 30 x height / 33000

=

1130 x height / 33000

therefore

the building's height = 33000/1130 x the amount of seconds between when you dropped the cracker and when you heard people swearing at you.

Sometimes maths can be fun. But usually only when it involves things that explode.

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A reasonable idea, but the distance of fall (ignoring friction) is $16t^2$, not $30t.$ –  Ross Millikan Oct 16 '12 at 16:24
    
@Ross. From my youthful pyrotechnic experience, a typical firecracker will reach its terminal velocity quite soon after being tossed off a building, so the linear term for velocity is actually a better answer (though some preliminary experiments should be done to determine the correct coefficient). –  Rick Decker Oct 16 '12 at 16:31
    
This answer, creative as it may be, requires a lot more information to compute the distance the firecracker would fall in a given amount of time (shape, size, and weight of the firecracker for starters). Not really an answer. –  robjohn Oct 16 '12 at 23:06

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