If a stone is tossed from the top of a 250 meter building, the height of the stone as a function of time is given by h(t) = -9.8t2 – 10t + 250, where t is in seconds, and height is in meters. After how many seconds will the stone hit the ground? Round to the nearest hundredth’s place; include units in your answer.
To find the time at which the stone hits the ground, we can set the height function, h(t), equal to zero and solve for t.
The given height function is: h(t) = -9.8t^2 - 10t + 250
Setting h(t) = 0, we have: -9.8t^2 - 10t + 250 = 0
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -9.8, b = -10, and c = 250. Plugging these values into the quadratic formula, we get:
t = (-(-10) ± √((-10)^2 - 4(-9.8)(250))) / (2(-9.8))
Simplifying further, we have:
t = (10 ± √(100 - (-9800))) / (-19.6)
t = (10 ± √(100 + 9800)) / (-19.6)
t = (10 ± √(9900)) / (-19.6)
Now, we need to evaluate the two possible values for t:
t1 = (10 + √(9900)) / (-19.6)
t2 = (10 - √(9900)) / (-19.6)
Rounding to the nearest hundredth's place, we have:
t1 ≈ -0.15 seconds
t2 ≈ 13.34 seconds
Since time cannot be negative in this context, we discard the negative value. Therefore, the stone will hit the ground after approximately 13.34 seconds.