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On Mars, if you hit a baseball, the height of the ball at time t would be modeled by the quadratic function h(t)=-1.85t2+20t+1, where t is seconds and h(t) is in metres.
A) When will the ball hit the ground?
B) How long will the ball be above 17m?

For A I got 5.41. But I'm sure that is wrong. 

 Jun 8, 2019
 #1
avatar+102372 
+2

OK....let's see

 

A)     -1.85t^2 + 20t + 1  = 0

 

Using the quadratic formula

 

        -20 ±√[ 20^2  - 4 (-1.85) (1) ]               -20 ±√  [ 400 + 7.4]

t =    _______________________     =       _______________

               2 (-1.85)                                               -3.7

 

t  =    -20 ±√ [407.4]

        ____________    =    10.86 sec     or  -.049  sec   (reject)

              -3.7 

 

Here's a graph  [ I used x instead of t ] : https://www.desmos.com/calculator/znjojz3ud3

 

cool cool cool    

 Jun 8, 2019
 #2
avatar+102372 
+2

For the second one we want to solve this

 

-1,85t^2 + 20t + 1  = 17       subtract 17 from both sides

 

-1.85t^2 + 20t  - 16   = 0

 

 

t  =    -20 ±√ [ 20^2 - 4(-1.85)(-16) ]                 -20 ±√ [ 281.6]

        __________________________  =        ____________  =

                     -3.7                                                    -3.7 

 

t = .87 sec      or t  = 9.94 sec

 

So...it will be above  17 m  for   [ 9.94 - .87 ]  ≈ 9.07 sec

 

 

cool cool cool

 Jun 9, 2019

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