Question 284087
A ball is thrown into the air with an initial upward velocity of 60ft/sec. Its height (h) in feet after 1 sec is given by the function h=-16t2(2= squared cant make a little 2)+60t +6
{{{h(t) = -16t^2 + 60t + 6}}}
Not for 1 second, but at t seconds
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a: after how many seconds will the ball hit the ground?
It hits the ground when h(t) = 0
{{{h(t) = -16t^2 + 60t + 6}}}
{{{0 = -16t^2 + 60t + 6}}}
{{{-8t^2 + 30t + 3 = 0}}}
*[invoke solve_quadratic_equation -8,30,3]
Use the + value, ~3.847 seconds
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b: what will the height be at if t= 3 seconds
h(3) = -16*9 + 60*3 + 6 = -144 + 180 + 6
h(3) = 42 feet
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i think for part be the answer is 42 ive plugged in 3 for t and got 42 i have no idea how to get part A