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avatar+50 

The equation $y = -16t^2 - 18t + 405$ describes the height (in feet) of a ball thrown downward at 18 feet per second from a height of 405 feet from the ground, as a function of time $t$, in seconds. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest tenth.

Thanks!

Don't know where to start.

 Apr 15, 2022
edited by Atlas  Apr 15, 2022
edited by Atlas  Apr 15, 2022
 #1
avatar+122390 
+2

When the ball hits the ground the  height   = 0

 

Multiplying the original equation through by -1   and  setting this to 0  we have

 

16t^2  + 18t  - 405  = 0

 

Use the Q Formula

 

 

 

t  =  (  -18  ± sqrt  [ 18^2  + 4*16 *405 ] )  /  [ 2 * 16 ]  =    [  -18 ±  162 ] / 32

 

Take the positive root

 

t =  [ -18 + 162 ]  /32   =    144 / 32   =    4.5 sec

 

cool cool cool

 Apr 15, 2022
 #2
avatar+50 
+1

Thanks!!

 Apr 15, 2022
 #3
avatar+122390 
+1

Welcome aboard, Atlas  !!!!!

 

 

cool cool cool

CPhill  Apr 15, 2022

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