Ted popped a baseball straight up into the air at a speed of 48 ft/s. The height of the ball after t seconds is given by the function h(t) = -16t2 + 48t + 3
When does the baseball hit the ground?
To find when the baseball hits the ground, we need to determine the value of t when the height of the ball is 0.
Setting h(t) equal to 0, we have:
-16t^2 + 48t + 3 = 0
Now, we can solve this quadratic equation for t. We can either use the quadratic formula or factor the equation, but in this case, the quadratic formula will be most straightforward:
t = (-b ± √(b^2 - 4ac)) / 2a
Substitute a = -16, b = 48, and c = 3 into the formula:
t = (-48 ± √(48^2 - 4(-16)(3))) / 2(-16)
t = (-48 ± √(2304 + 192)) / -32
t = (-48 ± √2496) / -32
Now, calculate the values of t:
t = (-48 + √2496) / -32
t = (-48 + 49.96) / -32
t = 1.96 / -32
t ≈ -0.06
t = (-48 - √2496) / -32
t = (-48 - 49.96) / -32
t = -97.96 / -32
t ≈ 3.06
Since time cannot be negative, the baseball hits the ground at approximately 3.06 seconds after it was thrown.