H(t) = -16t² + vt + s
Where -16 is half of the gravitational constant of almost 32 ft/sec (downward, thus negative),
v is the initial velocity (90), and
s is the starting height (160)
so we have:
H(t) = -16t² + 90t + 160
How long before it hits the ground? Solve for h(t) = 0:
0 = -16t² + 90t + 160
Divide both sides by -2:
0 = 8t² - 45t - 80
Quadratic equation:
t = [ -b ± √(b² - 4ac)] / (2a)
t = [ -(-45) ± √((-45)² - 4(8)(-80))] / (2(8))
t = [ 45 ± √(2025 + 2560)] / 16
t = [ 45 ± √(4585)] / 16
throwing out the negative time:
t = (45 + √4585) / 16
t ≈ 7.04 seconds
Answer:
Three fractions between 1/2 and 1 are 3/4, 4/5, and 5/6.
Answer:
im pretty sure you can do this with a protracter
Step-by-step explanation:
Answer:
The answer in simplified form would be 6x+12y.
Step-by-step explanation:
Answer:
y=-9
Step-by-step explanation:
slope intercept form: y=mx+b, m being the slope
y=6 times -3
since there isn't a y intercept, you can't really put it in standard form unless I missed a step.
y= -9
