A ball is kicked 4 feet above the ground with an initial vertical velocity of 55 feet per second. The function $h\left(t\right)=-16t^2+55t+4$ represents the height $h$ (in feet) of the ball after $t$ seconds. Using a graph, after how many seconds is the ball 30 feet above the ground? Round your answers to the nearest tenth. The ball is 30 feet above the ground after a second(s) and a second(s).

Guest Apr 7, 2020

#1**+1 **

16t^2+55t+4= 30 feet is the equation we need to solve this problem. In this case, we'll be solving for t.

16t^2+55t-26=0 since it does not factor, we'll have to do this out.

Since our answer cannot be negative, our solution is 0.42113353947124876.

Rounding to the nearest 10th we get **.4** as our final answer.

Hope it helps!

HELPMEEEEEEEEEEEEE Apr 7, 2020

#1**+1 **

Best Answer

16t^2+55t+4= 30 feet is the equation we need to solve this problem. In this case, we'll be solving for t.

16t^2+55t-26=0 since it does not factor, we'll have to do this out.

Since our answer cannot be negative, our solution is 0.42113353947124876.

Rounding to the nearest 10th we get **.4** as our final answer.

Hope it helps!

HELPMEEEEEEEEEEEEE Apr 7, 2020