We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948
then
let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is
23/15.948=1.44218 sec
substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329
determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon
terminal velocity ... greater speed ... acc is 10m/s/s
A) d. 10T
When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.
This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

which can be rewritten as

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

So, we get:

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.
B) 
The frequency of revolution of a particle in uniform circular motion is

where
f is the frequency
T is the period
We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:
T' = 10 T
Then its frequency of revolution will be:

The answer is negative 14