Answer:
H = Vy t - 1/2 g t^2 height of an object with an initial "vertical" velocity
at t sec after firing
Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s
H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m
Because the information cant be out of the investigation
In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.
This also means that the atomic number Z of the element (the atomic number is the number of protons in the nucleus) decreases by 2 units in the process, while the mass number A (the mass number is the sum of the number of protons and neutrons) decreases by 4 units.
With constant angular acceleration
, the disk achieves an angular velocity
at time
according to

and angular displacement
according to

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

b. Under constant acceleration, the average angular velocity is equivalent to

where
and
are the final and initial angular velocities, respectively. Then

c. After 1.00 s, the disk has instantaneous angular velocity

d. During the next 1.00 s, the disk will start moving with the angular velocity
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to

which would be equal to
