The angular velocity, ω=
2π/t; t = 24 hrs = 24 x 3600 seconds = 86400 s
ω = 7.27 x 10⁻⁵
v = ωr
= 7.27 x 10⁻⁵ x 3242.8 x 1.6 x 1000 (converting miles to meters)
= 377.2 m/s
I think it is acceleration
7. First write down all the known variables while separating the values for each direction:
x-direction:
vix = 20m/s
vfx = 20m/s
x = 39.2m
y-direction:
viy = 0m/s
ay = -9.8m/s^2
y = ?
Based on the knowns, the first step is to calculate the time of flight from the x-direction as it will be the same as value for the y-direction. Find the correct kinematic equation to do so:
x = (1/2)(vix+vfx)t
(39.2) = (1/2)(20+20)t
1.96s = t
Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the y-direction:
y = viy*t + (1/2)ay*t^2
y = (0)(1.96) + (1/2)(-9.8)(1.96)^2
y = -18.82m (Value is negative because gravity constant was negative. It is the height reference that from the top of the building down, which is why it is negative. The sign can be ignored for this question.)
8. First write down all the known variables while separating the values for each direction:
x-direction:
x = 12m
vfx = 0m/s
vix = ?
y-direction:
y1 = 1.2m
y2 = 0.6m
viy = 0m/s;
ay = -9.8m/s^2
First find time in the y-direction as it would be the same value for the x-direction.
(y2 - y1) = viy*t + (1/2)ay*t^2
(-0.6) = (0)t + (1/2)(-9.8)t^2
t = 0.35s
Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the x-direction:
x = (1/2)(vix+vfx)t
(12) = (1/2)(vix+(0))(0.35)
68.6m/s = vix
