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.

Guest Jan 4, 2015

#1**+10 **

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

h' =-32t + v_{0}

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

-32t + v_{0} = 0 → v_{0} = 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 v_{0} when t= √10 , we have ... v_{0} = 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

#1**+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 + v_{0}

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

-32t + v_{0} = 0 → v_{0} = 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 v_{0} when t= √10 , we have ... v_{0} = 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