Trying to find maximum height of projectile?

Question

I know that the mass shot from the cannon is 0.787 kg and the end of the cannon's barrel is at 6.9m and the initial velocity vi of the projectile has a horizontal component of 4.6m/s. The projectile rises to a height of delta y and strikes the ground at a horizontal distance delta x past the end of the cannon's barrel, and the acceleration of gravity is 9.8m/s^2. I have to find out the time it takes for the projectile to reach its maximum height, so if anyone could explain what I could do or if there are equations I could use to find the maximum height that would be helpful and appreciated.

Answer 1

Any number of different trajectories could produce the data you have given. (Note mass is irrelevant.) Do you know delta x, or is the solution to be nonnumeric in spite of all the numbers given?
If you know deltax, t = deltax/v(hor).
A general ballistic equation that accommodates a change in y (ie, y at target, yt, not = y at launch) is (from the ref.)
yt = range*tan(theta) - g*range^2*(1+tan(theta)^2)/(2*v0^2)
This can be permuted in various ways to solve for particular parameters that determine the trajectory, e.g.,
theta = arctan[(v0^2 +/- sqrt{v0^4 - g*(g*range^2+2*y*v^2)}) / (g*range)]
v0 = sqrt[0.5a*range^2*(1+tan(theta)^2) / (range*tan(theta)-yt)]
I hope this is some use.
P.S. The first answer confused hor. vel. for vert. It was too good to be true.

Answer 2

I'm not sure if this problem is solvable. Are you supposed to give a value for the answer or an expression?

If the inital vertical velocity was given, or initial velocity and angle were given so that you could find the initial vertical velocity, then it would be possible to find the max height using the equation. v^2-u^2=2ad and time taken using a=(v-u)/t

Also, if the distance where it lands was given, you could find the time taken to travel (v=d/t as horizontal velocity stays constant), and then find the other data, and find the max height/time.

Answer 3

Use v = u + at (final velocity = initial velocity + acceleration *time
At the maximum height v=0; so we have

0 = 4.6 - 9.8*t (negative sign because acceleration due to gravity os slowing the cannon ball down)

ie 4.6 = 9.8*t or t = 4.6/9.8 = 0.47 secs
You don't need to know the maximum height!
If you wanted to know it you could substitue relevant values into

v^2 = u^2 + 2*a*s (remembering that a is negative!) and solve for s