1) ![x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2](https://tex.z-dn.net/?f=x%3D%20v_%7B0x%7D%20t%20%3D%2023.4%20t%5C%5Cy%3Dy_0%20%2B%20v_%7B0y%7Dt-%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%3D%2012%2B21.8t%20-4.9t%5E2)
The initial data of the projectile are:
is the initial height
is the initial speed of the projectile, so its components along the x- and y- directions are
![v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s](https://tex.z-dn.net/?f=v_%7B0x%7D%20%3D%20v_0%20cos%20%5Ctheta%20%3D%20%2832%20m%2Fs%29%28cos%2043%5E%7B%5Ccirc%7D%29%3D23.4%20m%2Fs%5C%5Cv_%7B0y%7D%20%3D%20v_0%20sin%20%5Ctheta%20%3D%20%2832%20m%2Fs%29%28sin%2043%5E%7B%5Ccirc%7D%29%3D21.8%20m%2Fs)
The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:
![x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2](https://tex.z-dn.net/?f=x%3D%20v_%7B0x%7D%20t%20%3D%2023.4%20t%5C%5Cy%3Dy_0%20%2B%20v_%7B0y%7Dt-%5Cfrac%7B1%7D%7B2%7Dgt%5E2%20%3D%2012%2B21.8t%20-4.9t%5E2)
2) 4.94 s
To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:
![0=12+21.8t-4.9 t^2](https://tex.z-dn.net/?f=0%3D12%2B21.8t-4.9%20t%5E2)
Solving the equation with the formula, we have:
![t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}](https://tex.z-dn.net/?f=t_%7B1%2C2%7D%3D%5Cfrac%7B-21.8%5Cpm%20%5Csqrt%7B%2821.8%29%5E2-4%28-4.9%29%2812%29%7D%7D%7B2%28-4.9%29%7D)
which has two solutions:
t = -0.50 s
t = 4.94 s
Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is
t = 4.94 s
3) 115.6 m
To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:
![x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m](https://tex.z-dn.net/?f=x%3Dv_%7B0x%7Dt%3D%2823.4%20m%2Fs%29%284.94%20s%29%3D115.6%20m)
4) 2.22 s
The cannonball reaches its maximum height when the vertical velocity becaomes zero.
The vertical velocity at time t is given by
![v_y(t)= v_{0y} -gt](https://tex.z-dn.net/?f=v_y%28t%29%3D%20v_%7B0y%7D%20-gt)
where
g = 9.8 m/s^2 is the acceleration due to gravity
Substutiting
and solving for t, we find
![t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv_%7B0y%7D%7D%7Bg%7D%3D%5Cfrac%7B21.8%20m%2Fs%7D%7B9.8%20m%2Fs%5E2%7D%3D2.22%20s)
5) 36.2 m
The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:
![y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m](https://tex.z-dn.net/?f=y%28t%29%3Dy_0%20%2B%20v_%7B0y%7Dt-%5Cfrac%7B1%7D%7B2%7Dgt%5E2%3D12%2B%2821.8%29%282.22%29-%284.9%29%282.22%29%5E2%3D36.2%20m)