An object is thrown upward at a speed of 148 feet per second by a machine from a height of 20 feet off the ground. The height
h of the object after t seconds can be found using the equation
h=−16t2+148t+20
When will the height be 46 feet? (Note that there are TWO answers, one for moving up and one for moving down!)
Hint: Set h to 46 and solve for t.When will the object reach the ground?
Incorrect
seconds
−16t^2+148t+20 = 46
-16t^2+148t-26 = 0
t = (37±√1265)/8
To find when the height will be 46 feet, we can set the equation for height, h, equal to 46 and solve for t.
The equation for height, h, is given as h = -16t^2 + 148t + 20.
Setting h equal to 46, we have:
46 = -16t^2 + 148t + 20.
Now, we can rearrange the equation to set it equal to zero in order to solve for t:
-16t^2 + 148t + 20 - 46 = 0.
Simplifying, we get:
-16t^2 + 148t - 26 = 0.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = -16, b = 148, and c = -26.
Plugging in these values into the quadratic formula, we get:
t = (-148 ± √(148^2 - 4(-16)(-26))) / (2(-16)).
Simplifying further:
t = (-148 ± √(21904 - 1664)) / (-32).
t = (-148 ± √(20240)) / (-32).
Now, we can find the two solutions for t by taking both the positive and negative square roots:
t = (-148 + √(20240)) / (-32) and t = (-148 - √(20240)) / (-32).
Evaluating these values using a calculator, we get:
t ≈ 0.58 seconds and t ≈ 9.92 seconds.
Therefore, the height will be 46 feet at approximately 0.58 seconds and 9.92 seconds.