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A ball is thrown from an initial height of 2 meters with an initial upward velocity of 15 m/s. The balls height h (in meters) after t seconds is given by the following:h=2+15t-5t^2Find all values of t doe which the balls height is 7 meters. Round your answer to the nearest hundredth.

User PRATHAP S
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1 Answer

4 votes

Given:


h=2+15t-5t^2

Let the height of the ball is 7 meteres.

h=7,


7=2+15t-5t^2
5t^2-15t-2+7=0
5t^2-15t+5=0
t^2-3t+1=0
t=\frac{3\pm\sqrt[]{3^2-4(1)(1)}}{2(1)}
t=\frac{3\pm\sqrt[]{9-4}}{2}
t=\frac{3\pm\sqrt[]{5}}{2}
t=\frac{3+\sqrt[]{5}}{2},\frac{3-\sqrt[]{5}}{2}
t=(3+2.24)/(2),(3-2.24)/(2)
t=(5.24)/(2),(0.76)/(2)
t=2.62,0.38

The ball in the height of 7 meters at 0.38 seconds and 2.62 seconds.

User Hellopat
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