Aball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. the distance s (in feet) of the ball from the ground after t seconds is s equals 64 plus 48 t minus 16 t squared . (a) after how many seconds does the ball strike the ground? (b) after how many seconds will the ball pass the top of the building on its way down?

a) t = 4 secs

b) t = 3 secs

Step-by-step explanation:

h = 64 ft

Initial velocity = 48 ft/s

S = 64 + 48t - 16t^2

a) Set S = 0 and solve for t

0 = 64 + 48t - 16t^2

Divide through by 16

0 = 4 + 3t - t^2

t^2 - 3t - 4 = 0

t^2 - 4t + t - 4 = 0

t(t - 4) + 1(t - 4) = 0

(t + 1) (t - 4) = 0

t + 1 = 0 or t - 4 = 0

t = -1 or t = 4

t = {-1,4}

t = 4 seconds

b) Set S = 64 and solve for t

64 = 64 + 48t - 16t^2

64 - 64 = 48t - 16t^2

0 = 48t - 16t^2

Divide through by 16

0 = 3t - t^2

t^2 - 3t = 0

t(t - 3) = 0

t = 0 or t-3 = 0

t = 0 or t = 3

t = {0,3}

t = 3 secs

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

you estimate that the angle of elevation to the top of the 86th floor (the observatory) is 82 .the total height of the building is another 123 meters above the 86th floor.

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