A ball 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 dа ta:

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

a) t = 2 sec

b) 0.5s < t < 1.5s

Explanation:

Given dа ta:

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

s = 32t - 16t^2

0 = 32t-16t^2

0 = t (32-16t)

t = 0

32 = 16 t

t = 2 seconds

Answer: The ball strikes the ground again at t = 2 seconds

b) set given equation greater than 12 and solve for t:

s = 32 t - 16t^2

12 < 32t-16t^2

12 < 32t-16t^2

t^2 - 2t + 0.75 < 0

0.5s < t < 1.5s

Answer: The ball is above 12 feet from the ground in the time interval of 0.5s < t < 1.5s