What do I have wrong in here?
A stone is thrown vertically upward at a speed of 60.0m/s.
a) How long after being thrown is it 175m above ground?
So i did -
d = V1t + 1/2at^2
175 = 60t + 1/2 (10)t^2 (rearranged and divided it all by 5)
0 = t^2 + 12t - 35
And I can't get the factors.
We were asked to consider acceleration as 10.0m/s^2
Also..under what conditions will the quadratic be a perfect square?
I don't understand how to explain it.
Wouldn't v1 be 0?
To find the factors for the equation t^2 + 12t - 35 = 0, you can use the quadratic formula. The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the equation is in the form of at^2 + bt + c = 0, where a = 1, b = 12, and c = -35. Now you can substitute these values into the quadratic formula:
t = (-12 ± √(12^2 - 4 * 1 * -35)) / (2 * 1)
Simplifying this expression:
t = (-12 ± √(144 + 140)) / 2
t = (-12 ± √284) / 2
t = (-12 ± √(4 * 71)) / 2
t = (-12 ± 2√71) / 2
You can simplify it further:
t = -6 ± √71
So the values of t are t = -6 + √71 and t = -6 - √71. However, since time cannot be negative in this context, you can discard the negative value.
Therefore, the answer to the question "How long after being thrown is it 175 m above the ground?" is t = -6 + √71 (approximately 3.129 seconds).
Please note that in physics problems like this, you need to consider the direction of motion. Since the stone is thrown vertically upward, it reaches its highest point, then falls back down. So keep in mind that you may need to consider both the positive and negative values when dealing with time.