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The Stratosphere Tower in Las Vegas is 921 feet tall and has
a “needle” at its top that extends even higher into the air. A thrill ride called the
Big Shot catapults riders 160 feet up the needle and then lets them fall back to
the launching pad.

 

a.
The height h (in feet) of  a rider on the Big Shot can be modeled by h= º16t2+v0t+ 921 where tis the elapsed
time (in seconds) after launch and v 0 is the initial vertical velocity (in feet per second). Find v0 using the fact that the maximum value of h is 921 + 160 = 1081 feet.
 Jan 4, 2015

Best Answer 

 #1
avatar+105411 
+10

 

 

Using Calculus, we can find the derivatve of the function.... which is the velocity at any time t..so we have

h' =-32t + v0

And at the max ht, this is = 0, so we have

-32t + v0 = 0     →  v0 = 32t  ...and, substituting this back into the position function at the time of the max ht, we can find the time it takes to reach that point...so we have

1081 = -16t^2 + (32t)t + 921    simplify

1081 = -16t^2 + 32t^2 + 921

160 = 16t^2    divide both sides by 32

10 = t^2

t = √10 sec  ..and this is the time it takes to reach the max ht

So, using the first derivative to solve for v0 when t= √10  , we have ... v0  = 32(√10)ft/s

And our original position function becomes

h = -16t^2 + 32√10t + 921

Here's the graph of the function.....https://www.desmos.com/calculator/42nlumsshk

--------------------------------------------------------------------------------------------------------------

As I'm not great at Physics, could someone else (Alan or Melody??) check my answer  ???

 

 Jan 5, 2015
 #1
avatar+105411 
+10
Best Answer

 

 

Using Calculus, we can find the derivatve of the function.... which is the velocity at any time t..so we have

h' =-32t + v0

And at the max ht, this is = 0, so we have

-32t + v0 = 0     →  v0 = 32t  ...and, substituting this back into the position function at the time of the max ht, we can find the time it takes to reach that point...so we have

1081 = -16t^2 + (32t)t + 921    simplify

1081 = -16t^2 + 32t^2 + 921

160 = 16t^2    divide both sides by 32

10 = t^2

t = √10 sec  ..and this is the time it takes to reach the max ht

So, using the first derivative to solve for v0 when t= √10  , we have ... v0  = 32(√10)ft/s

And our original position function becomes

h = -16t^2 + 32√10t + 921

Here's the graph of the function.....https://www.desmos.com/calculator/42nlumsshk

--------------------------------------------------------------------------------------------------------------

As I'm not great at Physics, could someone else (Alan or Melody??) check my answer  ???

 

CPhill Jan 5, 2015
 #2
avatar+106046 
+5

Chris has asked me to look at this one and YES Chris I got the same answer.

I did it almost the same the only minor difference was that I repositioned the origin from the ground to the bottom of the needle.  I doubt that it made it any easier though.

 Jan 5, 2015
 #3
avatar+105411 
0

Thanks, Melody....

 

 Jan 5, 2015
 #4
avatar+28250 
+5

I agree that Chris's analysis is correct (apart from the bit that says "divide both sides by 32"  - this should say "divide both sides by 16"!!)

.

 Jan 5, 2015

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