A ball thrown vertically from the top of a building 96m at initial velocity 80m/sec.the distance 's' of the ball of the ground after 't' second is s=96+80t-4.5t^2. After how many seconds does the ball strike the ground...?
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To find out after how many seconds the ball strikes the ground, we need to find the time when the ball's height (s) becomes zero.
The equation for the height of the ball (s) is given as s = 96 + 80t - 4.5t^2.
Setting s equal to zero, we have:
0 = 96 + 80t - 4.5t^2
Now, let's solve this quadratic equation to find the values of t.
Rearranging the equation, we have:
4.5t^2 - 80t - 96 = 0
To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 4.5, b = -80, and c = -96.
Plugging these values into the quadratic formula, we get:
t = (-(-80) ± √((-80)^2 - 4 * 4.5 * (-96))) / (2 * 4.5)
Simplifying further:
t = (80 ± √(6400 + 1728)) / 9
Now, let's calculate the values of t:
t = (80 ± √(8128)) / 9
Using a calculator, the square root of 8128 is approximately 90.14.
Substituting this value into the equation:
t = (80 ± 90.14) / 9
Now, let's calculate both possibilities:
t1 = (80 + 90.14) / 9 ≈ 17.8 seconds
t2 = (80 - 90.14) / 9 ≈ -1.1 seconds
Since time cannot be negative in this context, we discard the negative value of t.
Therefore, the ball strikes the ground approximately after 17.8 seconds.