Aball is thrown vertically upward with an initial velocity of 32 feet per second. the distance s (feet) of the ball from the ground after t seconds is s equals 32t-16t^2. at what time t will the ball strike the ground? for what time t is the ball more than 12 feet above the ground

a) t =2 sec

b)  t = 0.5 sec

Explanation:

Given data:

a) set s=given equation to zero and solve for t:

s = 32t - 16t^2

0 = 32t-16t^2

32t = 16t^2

32 = 16 t

t = 2 seconds

b) set given equation equal to 12 and solve for t:

s = 32 t -16t^2

12 = 32t-16t^2

12 = 32t-16t^2

t^2 -2t + 0.75 = 0

t = 1.5, 0.5

hence t = 0.5 sec is right answer

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