A ball is thrown vertically upward with an initial velocity of 96 feet per second.The distance s (in feet) of the ball from the ground after t seconds is&nbsp;s(t)=96t−16t2.&nbsp;(a) At what time t will the ball strike the ground?&nbsp;(b) For what time t is the ball more than 128 feet above the ground?

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A ball is thrown vertically upward with an initial velocity of 96 feet per second.The distance s (in feet) of the ball from the ground after t seconds is&amp;nbsp;s(t)=96t−16t2.&amp;nbsp;(a) At what time t will the ball strike the ground?&amp;nbsp;(b) For what time t is the ball more than 128 feet above the ground?

Bukvald5z · Accepted Answer

We have a ball is thrown vertically upward with an initial velocity of 96 feet per second.The distance s (in feet) of the ball from the ground after t seconds is&amp;nbsp;s(t)=96t−16t2.&amp;nbsp;a) we have&amp;nbsp;s(t)=96(t)−16t2=0&amp;nbsp;since the distance from the ground is s(t) so we have&amp;nbsp;s(t)=96t−16t2=0&amp;nbsp;t(96−16t)=0&amp;nbsp;t=0&amp;nbsp;or&amp;nbsp;t=6&amp;nbsp;the ball is on the ground at t=0 seconds and t=6 seconds when the ball is thrown and when the ball lands.&amp;nbsp;So, the ball strikes the ground after 6 seconds.

Ella Williams · Accepted Answer

b) when is&amp;nbsp;s=128&amp;nbsp;so putting this value in the function then we have&amp;nbsp;128=96t−16t2&amp;nbsp;t2−6t+8=0&amp;nbsp;(t−4)(t−2)=0&amp;nbsp;that is between 2 seconds and 4 seconds.

nick1337 · Accepted Answer

(a) To find the time at which the ball strikes the ground, we need to determine when the height, represented by s(t), becomes zero. In this case, we can set up the equation:0=96t−16t2To solve this quadratic equation, we can factor it or use the quadratic formula. Let&#039;s use the quadratic formula:t=−b±b2−4ac2aFor our equation 0=96t−16t2, we have a=−16, b=96, and c=0. Substituting these values into the quadratic formula:t=−96±962−4(−16)(0)2(−16)Simplifying further:t=−96±9216−32t=−96±96−32We have two solutions:t1=−96+96−32=0t2=−96−96−32=6Therefore, the ball will strike the ground at t=6 seconds.(b) To find the time when the ball is more than 128 feet above the ground, we need to determine when the height, represented by s(t), is greater than 128. We can set up the inequality:128&amp;lt;96t−16t2Rearranging the inequality:16t2−96t+128&amp;lt;0Dividing all terms by 16 to simplify:t2−6t+8&amp;lt;0Now we can factor this quadratic inequality:(t−2)(t−4)&amp;lt;0To find when the inequality is true, we consider the sign changes of the quadratic expression. From the factored form, we see that the inequality is true when 2&amp;lt;t&amp;lt;4. Therefore, the ball is more than 128 feet above the ground for 2&amp;lt;t&amp;lt;4 seconds.

Don Sumner · Accepted Answer

Result:(a) The ball strikes the ground at t=0 and t=6 seconds.(b) The ball is more than 128 feet above the ground for 2&amp;lt;t&amp;lt;4 seconds.Solution:(a) At what time t will the ball strike the ground?To find the time t when the ball strikes the ground, we need to determine when the distance s(t) is equal to zero. Substituting the given equation s(t)=96t−16t2 into the condition s(t)=0, we have:96t−16t2=0Factoring out 16t from the equation, we get:16t(6−t)=0Now, we can set each factor equal to zero and solve for t:16t=0⇒t=06−t=0⇒t=6Thus, the ball strikes the ground at t=0 and t=6 seconds.(b) For what time t is the ball more than 128 feet above the ground?To determine the time t when the ball is more than 128 feet above the ground, we need to find when s(t)&amp;gt;128. Substituting the given equation s(t)=96t−16t2 into the inequality s(t)&amp;gt;128, we have:96t−16t2&amp;gt;128Rearranging the terms to form a quadratic inequality, we get:16t2−96t+128&amp;lt;0Dividing the inequality by 16 to simplify, we have:t2−6t+8&amp;lt;0To solve this quadratic inequality, we can factor it as:(t−2)(t−4)&amp;lt;0Now, we can use the sign analysis method to determine the range of t values that satisfy the inequality:t−2t−4t2−6t+8t&amp;lt;2−−+2&amp;lt;t&amp;lt;4+−−t&amp;gt;4+++From the sign analysis, we observe that the quadratic inequality is satisfied when 2&amp;lt;t&amp;lt;4. Therefore, the ball is more than 128 feet above the ground for 2&amp;lt;t&amp;lt;4 seconds.

A ball is thrown vertically upward with an initial velocity of 96 feet

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