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=-16t2+46t+6. after how many second will the ball hit the ground
To find the time at which the ball hits the ground, we need to determine when the height of the ball, h, equals zero.
Setting h = 0 in the equation h = -16t^2 + 46t + 6, we get:
0 = -16t^2 + 46t + 6
Now, we need to solve this quadratic equation. One way to do this is by factoring, but it might not be easy to find two numbers that multiply to -16*6=-96 and add up to 46.
Alternatively, we can make use of the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by the formula:
x = (-b ± √(b^2-4ac)) / (2a)
In this case, a = -16, b = 46, and c = 6. Substituting these values into the quadratic formula, we find:
t = (-46 ± √(46^2 - 4(-16)(6))) / (2(-16))
Simplifying further:
t = (-46 ± √(2116 + 384)) / (-32)
t = (-46 ± √2500) / (-32)
Taking the square root of 2500 gives:
t = (-46 ± 50) / (-32)
Now, we can solve for both possible values of t:
t1 = (-46 + 50) / (-32) = 4 / (-32) = -1/8
t2 = (-46 - 50) / (-32) = -96 / (-32) = 3
Since time cannot be negative in this context, we discard t1 = -1/8.
Therefore, the ball will hit the ground after 3 seconds.