A ball is thrown vertically upward from the ground. The distance s in t seconds is given by s = 4t^2 + t – 3.
A ball is thrown vertically upward from the ground. The distance s in t seconds is given by s = 4t2 + t – 3. After how many seconds does the ball come to rest?
A) 2
B) 1
C) 3/4
D) 4/3
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Correct option is (C) \(\frac{3}{4}\)
For ball come to rest, S should be zero.
\(\therefore\) \(4t^2+t-3=0\) \((\because s=4t^2+t-3)\)
\(\Rightarrow t=\frac{-1\pm\sqrt{1^2-4\times4\times-3}}{2\times4}\)
\(=\frac{-1\pm\sqrt{1+48}}{8}\)
\(=\frac{-1\pm\sqrt{49}}{8}\)
\(=\frac{-1\pm7}{8}\)
\(=\frac{-1+7}{8}\) or \(\frac{-1-7}{8}\)
\(=\frac{6}{8}\) or \(\frac{-8}{8}\)
\(=\frac34\;or\;-1\)
\(\because\) Time never be negative
\(\therefore\) t = \(\frac34\) second
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