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A ball is thrown vertically upward with an initial velocity of96 feet per second. The distance s (in feet) of the ball from theground after t seconds is s(t)= 96t - 16t2

A) At what time t will the ball strike the ground?
B) For what time t is the ball more than 128 feet above theground?
Step by step on how you did this recieve life saver pointsthanks for the help

1 Answer

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Final answer:

A) The ball will strike the ground after 6 seconds. B) The ball will be more than 128 feet above the ground when t is between 2 and 4 seconds.

Step-by-step explanation:

A) To find the time when the ball strikes the ground, we need to set the equation s(t) = 0 and solve for t. The equation s(t) = 96t - 16t^2 represents the distance of the ball from the ground after t seconds. Setting s(t) = 0, we have 0 = 96t - 16t^2. Rearranging the equation gives us 16t^2 - 96t = 0. Factoring out 16t, we get 16t(t - 6) = 0. This equation is true when t = 0 or t - 6 = 0. Since time cannot be negative, we discard t = 0. Therefore, the ball will strike the ground after 6 seconds.

B) To find the time when the ball is more than 128 feet above the ground, we need to set the equation s(t) > 128 and solve for t. Setting s(t) = 96t - 16t^2 > 128, we have 16t^2 - 96t + 128 < 0. Factoring out 16, we get 16(t^2 - 6t + 8) < 0. Factoring the quadratic equation gives us 16(t - 2)(t - 4) < 0. Using the sign chart method, we find that the inequality is true when t is in the interval (2, 4). Therefore, the ball will be more than 128 feet above the ground when t is between 2 and 4 seconds.

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