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# A ball is thrown.

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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).

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!

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