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

Beginner2022-01-10Added 33 answers

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

$s\left(t\right)=96t-16{t}^{2}=0$

$t(96-16t)=0$

$t=0$ 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

Beginner2022-01-11Added 28 answers

b) when is $s=128$ so putting this value in the function then we have

$128=96t-16{t}^{2}$

${t}^{2}-6t+8=0$

$(t-4)(t-2)=0$

that is between 2 seconds and 4 seconds.

nick1337

Expert2023-06-11Added 777 answers

Don Sumner

Skilled2023-06-11Added 184 answers

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