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 h equals negative 16 t squared plus 46 t plus 6. After how many seconds will the ball hit the ground?

Answer 1

To find out when the ball hits the ground, we need to determine when the height (h) is equal to 0.

Setting h equal to 0 in the given equation:
0 = -16t^2 + 46t + 6

Now, we have a quadratic equation that we can solve for t. We can either factor or use the quadratic formula.

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a

Here, a = -16, b = 46, and c = 6.

Plugging in the values:
t = (-46 ± √(46^2 - 4(-16)(6))) / 2(-16)
t = (-46 ± √(2116 + 384)) / -32
t = (-46 ± √2500) / -32
t = (-46 ± 50) / -32

Two possible solutions will be obtained:
1) t = (-46 + 50) / -32 = 4 / -32 = -1/8
2) t = (-46 - 50) / -32 = -96 / -32 = 3

Since time cannot be negative, the ball will hit the ground after 3 seconds.