+0

0
217
2
+156

Apr 24, 2018

#1
+3

2)

Solve for t:
41 = -16 t^2 + 50 t + 6

41 = -16 t^2 + 50 t + 6 is equivalent to -16 t^2 + 50 t + 6 = 41:
-16 t^2 + 50 t + 6 = 41

Divide both sides by -16:
t^2 - (25 t)/8 - 3/8 = -41/16

t^2 - (25 t)/8 = -35/16

t^2 - (25 t)/8 + 625/256 = 65/256

Write the left hand side as a square:
(t - 25/16)^2 = 65/256

Take the square root of both sides:
t - 25/16 = sqrt(65)/16 or t - 25/16 = -sqrt(65)/16

t = 25/16 + sqrt(65)/16 or t - 25/16 = -sqrt(65)/16

t = 25/16 + sqrt(65)/16 =~2.07 sec.                       or t = 25/16 - sqrt(65)/16(Discard this one)

Apr 24, 2018
edited by Guest  Apr 24, 2018
#2
+101322
+3

First one :  Either method is correct

The first student is evaluating  [25^(1/2)]^3    while the second student is evaluating (25^3)^(1/2)

Both lead to the same result

Second one

h  =  - 16t^2 + 50t + 6....we want to find the time when the height of the ball = 41 feet

So.... we can solve this

-16t^2  + 50t + 6   =  41    subtract 41 from both sides

-16t^2 + 50t - 35 =  0     multiply through by -1

16t^2 - 50t + 35  =   0

Using  the quadratic formula   where a = 16, b  = -50  , c  = 35

x  = ( - b  ±√[ b^2 - 4ac] )  / 2a  =   ( 50 ±√ [ (-50)^2 - 4(16)(35) ] ) / (2*16)  =

(50 ±√[ 260]) / 32

Taking  x  =  ( 50 + √[260] ) / 32

x  = 2.07 sec

Apr 24, 2018