a ball is thrown downward from the too of 110 ft bldg with an initial velocity of 14ft per second. the height of ball h after t seconds is given by the equation h = -16t^2-14t +110.
how long after ball is thrown will it hit the ground
when it is at ground level , h = 0
16t^2 + 14t - 110=0
8t^2 + 7t - 55 = 0
t = (-7 ± √1809)/16 = 2.22 or a negative
so it took 2.22 seconds before it hit the ground
To find out how long it takes for the ball to hit the ground, we need to find the time when the height (h) is equal to zero.
The equation for the height of the ball is given by h = -16t^2 - 14t + 110.
Setting h to zero, we get:
0 = -16t^2 - 14t + 110
Now, we have a quadratic equation that we can solve to find the value(s) of t when h is equal to zero.
To solve this quadratic equation, we can use any method, such as factoring, completing the square, or using the quadratic formula.
In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Using the equation 0 = -16t^2 - 14t + 110, we identify the coefficients:
a = -16, b = -14, c = 110
Plugging these values into the quadratic formula, we get:
t = (-(–14) ± √((-14)^2 - 4(-16)(110))) / (2(-16))
Simplifying further:
t = (14 ± √(196 + 7040)) / (-32)
t = (14 ± √(7236)) / (-32)
t = (14 ± 85.07) / (-32)
Now, we have two possible solutions for t:
1. t = (14 - 85.07) / -32 ≈ -2.03
2. t = (14 + 85.07) / -32 ≈ -3.03
Since time cannot be negative in this context, we discard the negative value.
Therefore, the ball will hit the ground approximately 3.03 seconds after it is thrown.