A skyscraper...

Question

A skyscraper has an outdoor observation deck that is 1524 feet high. An object is thrown upward with an initial velocity of 20 feet per second off the edge of the observation deck. The height h of the object in feet after t seconds is given by the function h(t)=-16t^2+20+1524. How long will it take for the object to be 1348 feet above the ground?

Answer 1

First, let this problem be motivation for you to memorize the general equation of motion for an object under constant acceleration:

 

d_t = d₀ + v₀t + (1/2)at², where d₀ is the initial position, a is the acceleration (usually gravity), and v₀t is the initial velocity.

 

What's hanging you up is that if you don't know the general equation, you won't see that the equation of motion you were given neglects to put "t" after "20." 20 feet per second is positive as you're tossing the coin upwards (to clear a fence or something). Note the acceleration is negative -- or downward -- and that its value is -16. This is one-half of 32 ft/s², which is gravity.

 

Once you've corrected the equation of motion, solve for t when the height is 1348 feet:

 

1348 = -16*t² + 20*t + 1524.

 

I like my squared terms positive, so move all to the left to get

 

16t² - 20t - 176 = 0. Simplify this by dividing by 4 to get

 

4t² - 5t - 44 = 0.

 

Play with it to factor it. (Use  "(4t + __)(t - __) = 0" to start.) Only the right term makes sense since it gives you a positive value for t.

 

Note this object falls so that air resistance is negligible. It's not a penny!

 

 

Eric Moline

McMinnville OR