Question

Two-dimensional motion HELP

Answer 1

hint:
the time \(\Delta t \) of flight for each shell, is given by this formula:
\[\Large \Delta t = \frac{{2{V_0}\sin \theta }}{g}\]
where:
\(V_0\) is the magnitude of the velocity of such shell, and \(\theta\) is its elevation.
Of course, \(g=32\) \( feet/sec^2\) is the gravity

Answer 2

@Stewart.S
Be wary of the formula given by @Michele_Laino as the question has the cannon lower than the target.
The given equation can be derived by assuming that the time it takes a projectile to get to its maximum height is the same as the time it takes to fall the SAME distance. In this example the distance travelled upwards is greater than the distance falling.

Here is how to derive the equation.
Start from one of the equations of constant acceleration vf = vi +at
For falling the initial velocity vi = 0, the final velocity is v0 sin theta and the acceleration is g

Putting this into the constant acceleration equation gives t = v0 sin theta / g

Now double the time to get the time to go up plus the time to go down and you get @Michele_Laino's equation.

Answer 3

For projectile type problems you need to use the following equations:

Answer 4

This problem has two unknowns for each of the projectile motions.
1 The initial velocity Vo
2 The time of flight t
So you need two equations to find these unknowns.
One is found bu considering horizontal motion with vix = V0 cos theta.
x = vix will give x = V0 cos theta t and you are given x and theta.

The other equation is one of the four from the ones listed above.and I suggest y = vi t + a t^2 / 2 is the one to go for.
y is the height difference between cannon and target, viy is V0 cos theta and a = -g as up is taken to be the positive direction.
You will get a quadratic in t and you must choose the positive solution.

Answer 5

more precisely, we can write these equations:

\[\Large\left\{ \begin{gathered}
\ddot z = - g \hfill \\
\ddot x = 0 \hfill \\
\end{gathered} \right.\]