3) Amy hit a softball into the air at a speed of 39 ft/s. The height of the ball after t seconds is given by the function h(t) = -16t2 + 39t + 2
When does the softball hit the ground?
To find when the softball hits the ground, we need to find the time when the height is equal to 0.
So, set h(t) = -16t^2 + 39t + 2 equal to 0:
-16t^2 + 39t + 2 = 0
This is a quadratic equation which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For this equation, a = -16, b = 39, and c = 2. Plugging these values into the formula:
t = (-39 ± √(39^2 - 4(-16)(2))) / 2(-16)
t = (-39 ± √(1521 + 128)) / -32
t = (-39 ± √1649) / -32
Now, calculate the value under the square root:
√1649 ≈ 40.61
t = (-39 ± 40.61) / -32
This gives two possible values for t:
t = (-39 + 40.61) / -32 ≈ 1.61 / -32 ≈ -0.05
t = (-39 - 40.61) / -32 ≈ -79.61 / -32 ≈ 2.49
Since time cannot be negative in this context, the answer is t ≈ 2.49 seconds.
Therefore, the softball hits the ground after approximately 2.49 seconds.