1)
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
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:
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:
Solving the equation with the formula, we have:
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:
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
where
g = 9.8 m/s^2 is the acceleration due to gravity
Substutiting and solving for t, we find
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: