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a ball is thrown from an initial height of 4 feet with an initial upward velocity of 35 ft per secondthe balls height H in feet after T seconds is given by the following h=4+35t-16t^2find all values of t for which is the balls height is 22 ft . round to the nearest hundredth

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Solution

For this case we can do the following:


4+35t-16t^2=22

And for this case we can rewrite this expression like this:


-16t^2+35t-18=0

And using the quadratic formula we got:


t=\frac{-35\pm\sqrt[]{(-35)^2-4(-16)(-18)}}{2\cdot(-16)}=\frac{-35\pm\sqrt[]{73}}{-32}

And the two solutions for this case are:

t= 1.36 s

t= 0.83 s

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