The height (h) of the ball after t seconds is given by the quadratic function h(t) = -16t^2 + 46t.
To find when the ball hits the ground, we need to find the value of t when h(t) = 0.
Setting -16t^2 + 46t = 0, we can factor out -2t to get -2t(8t - 23) = 0.
Setting each factor equal to zero gives us -2t = 0 and 8t - 23 = 0.
From -2t = 0, we get t = 0.
From 8t - 23 = 0, we can solve for t by adding 23 to both sides and dividing both sides by 8, giving us t = 23/8 = 2.875.
So, after 2.875 seconds, the ball will hit the ground.
However, this answer is not one of the given options. This might be a typo or mistake in the question.
A ball is thrown into the air with an initial upward velocity of 46 ft/s. Its height (h) in feet after t seconds is given by the function . After how many seconds will the ball hit the ground? (1 point) Responses 3 3 4 4 5 5 6
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