An object is launched upward from the ground with an initial vertical velocity of 40 feet per second. After how many seconds does the object reach a height of 25 feet?
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To find out how many seconds it takes for the object to reach a height of 25 feet, we can use the equation of motion:
s = ut + (1/2)gt^2
Where:
s = height (25 feet in this case)
u = initial velocity (40 feet per second)
t = time (unknown)
g = acceleration due to gravity (approximately -32 feet per second squared)
We can rearrange the equation to solve for time (t):
25 = 40t + (1/2)(-32)t^2
Next, we can simplify the equation:
25 = 40t - 16t^2
Let's move all the terms to one side of the equation to form a quadratic equation:
16t^2 - 40t + 25 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -40, and c = 25.
Plugging in the values, we can calculate:
t = (-(-40) ± sqrt((-40)^2 - 4(16)(25))) / (2(16))
Simplifying further:
t = (40 ± sqrt(1600 - 1600)) / 32
Since the term inside the square root is zero, the equation simplifies even further:
t = 40 / 32
t = 1.25 seconds
Therefore, it takes approximately 1.25 seconds for the object to reach a height of 25 feet.