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An object has been launched vertically upward from a building. The height of the object relate to time can be found by h(t)= -16t^2 +48t+960. (pls use this when solving the problem thanks)

 

A. What was the speed when the object was launched?

 

B. How tall was the building?

 

C. What is the maximum height the object could reach?

 

D. When will the object reach its maximum height?

 

 

E. When will the object hit the ground?

 

Thanks!

 Jun 1, 2021
 #1
avatar+121056 
+2

A   = 48 ft/ s

 

B =   960 ft

 

C   Find  the  t coordinate   of  the  vertex  =  - 48 / (2 * -16)   = 48/32  =  3/2  sec

Plug this  back into the  function

-16(3/2)^2   +  48(3/2)  + 960   =      996 ft = maax height

 

D  ( from C )  =   3/2  sec

 

E  The  object  will hit to  ground when  h  =  0....so

 

-16t^2  +  48t  +  960   =   0              divide through by  -16

 

t^2  - 3t   -  160   =  0         complete the square on t

 

t^2  -3t +  9/4   = 160 +  9/4

 

(t  - 3/2)^2   = 649/4               take both  roots and  we  have  that

 

t  = sqrt (694)/2 + 3/2  = 14.67 sec            or      t  =  -sqrt (694/)/2  +3/2  = -11.67 sec

 

Take the positive  answer  =    14.67  sec  

 

 

cool cool cool

 Jun 1, 2021
 #2
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Thanks CPhill!!! I made it way more complicated then it needed to be. Thanks!

 Jun 1, 2021
 #3
avatar+121056 
0

You're welcome !!!

 

 

cool cool cool

CPhill  Jun 1, 2021

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