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

Expert

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

Expert

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.

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