1) [tex]x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2[/tex]
The initial data of the projectile are:
[tex]y_0 = 12 m[/tex] is the initial height
[tex]v_0 = 32 m/s[/tex] is the initial speed of the projectile, so its components along the x- and y- directions are
[tex]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[/tex]
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:
[tex]x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2[/tex]
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:
[tex]0=12+21.8t-4.9 t^2[/tex]
Solving the equation with the formula, we have:
[tex]t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}[/tex]
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:
[tex]x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m[/tex]
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
[tex]v_y(t)= v_{0y} -gt[/tex]
where
g = 9.8 m/s^2 is the acceleration due to gravity
Substutiting [tex]v_y(t)=0[/tex] and solving for t, we find
[tex]t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s[/tex]
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:
[tex]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[/tex]
