Sapewa

2022-01-09

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 $s\left(t\right)=96t-16{t}^{2}$

(a) At what time t will the ball strike the ground?

(b) For what time t is the ball more than 128 feet above the ground?

Bukvald5z

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 $s\left(t\right)=96t-16{t}^{2}$

a) we have $s\left(t\right)=96\left(t\right)-16{t}^{2}=0$
since the distance from the ground is s(t) so we have
$s\left(t\right)=96t-16{t}^{2}=0$
$t\left(96-16t\right)=0$
$t=0$ or $t=6$
the ball is on the ground at t=0 seconds and t=6 seconds when the ball is thrown and when the ball lands.
So, the ball strikes the ground after 6 seconds.

Ella Williams

b) when is $s=128$ so putting this value in the function then we have
$128=96t-16{t}^{2}$
${t}^{2}-6t+8=0$
$\left(t-4\right)\left(t-2\right)=0$
that is between 2 seconds and 4 seconds.

nick1337

(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:
$0=96t-16{t}^{2}$
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:
$t=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
For our equation $0=96t-16{t}^{2}$, we have $a=-16$, $b=96$, and $c=0$. Substituting these values into the quadratic formula:
$t=\frac{-96±\sqrt{{96}^{2}-4\left(-16\right)\left(0\right)}}{2\left(-16\right)}$
Simplifying further:
$t=\frac{-96±\sqrt{9216}}{-32}$
$t=\frac{-96±96}{-32}$
We have two solutions:
${t}_{1}=\frac{-96+96}{-32}=0$
${t}_{2}=\frac{-96-96}{-32}=6$
Therefore, 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:
$128<96t-16{t}^{2}$
Rearranging the inequality:
$16{t}^{2}-96t+128<0$
Dividing all terms by 16 to simplify:
${t}^{2}-6t+8<0$
Now we can factor this quadratic inequality:
$\left(t-2\right)\left(t-4\right)<0$
To 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. Therefore, the ball is more than 128 feet above the ground for $2 seconds.

Don Sumner

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 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 $s\left(t\right)$ is equal to zero. Substituting the given equation $s\left(t\right)=96t-16{t}^{2}$ into the condition $s\left(t\right)=0$, we have:
$96t-16{t}^{2}=0$
Factoring out $16t$ from the equation, we get:
$16t\left(6-t\right)=0$
Now, we can set each factor equal to zero and solve for $t$:
$16t=0\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}t=0$
$6-t=0\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}t=6$
Thus, 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 $s\left(t\right)>128$. Substituting the given equation $s\left(t\right)=96t-16{t}^{2}$ into the inequality $s\left(t\right)>128$, we have:
$96t-16{t}^{2}>128$
Rearranging the terms to form a quadratic inequality, we get:
$16{t}^{2}-96t+128<0$
Dividing the inequality by 16 to simplify, we have:
${t}^{2}-6t+8<0$
To solve this quadratic inequality, we can factor it as:
$\left(t-2\right)\left(t-4\right)<0$
Now, we can use the sign analysis method to determine the range of $t$ values that satisfy the inequality:
$\begin{array}{cccc}\hfill \hfill & \hfill t-2\hfill & \hfill t-4\hfill & \hfill {t}^{2}-6t+8\hfill \\ \hfill t<2\hfill & \hfill -\hfill & \hfill -\hfill & \hfill +\hfill \\ \hfill 24\hfill & \hfill +\hfill & \hfill +\hfill & \hfill +\hfill \end{array}$
From the sign analysis, we observe that the quadratic inequality is satisfied when $2. Therefore, the ball is more than 128 feet above the ground for $2 seconds.

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