can anyone help me with projectile motion?

Height at any time t = y(t) = h + Uy t - 1/2 g t^2

Distance at any time t = x(t) = Ux t

g = 9.810 m/s^2 (ft/s^2)

Launch height above ground H = 1.000 meters (feet)

Impact (target) elevation y(T) = 0.000 m (ft)

h = H - y(T) = 1.000 m (ft)

Launch speed U = 56.000 mps (fps)

Launch angle theta = 69.000 degrees

Uy = U sin(theta) = 52.281 mps (fps)

Ux = U cos(theta) = 20.069 mps (fps)

To find total flight time T, we solve the quadratic: 0 = 1.000 + 52.281 T - 4.905 T^2

Total Flight Time seconds

T = 10.678

T' = -0.019

Max Range to Impact meters (feet)

x(T) = Ux T = 214.287 ANS <====================

For nearly any projectile motion, breaking the velocity vector into it's component vectors is usually the simplest approach.

First let's find the horizontal component. There should be no deceleration in the horizontal direction if we assume air resistance as negligible. So then this component is just:

v = 56*cos69 = 20.1 m/s

This means that the cannonball will travel horizontally at a constant speed of 20.1 m/s.

Now let's find the vertical component. This one is a little more difficult because in the vertical direction, we have an acceleration in the downward direction due to gravity. To include this, we just subtract the acceleration of gravity, times the amount of time that has passed:

v = 56*sin(69) - 9.81t = 52.3 - 9.81t

The object here is to find out how far the projectile has traveled. To do that we need to know how long the projectile is in the air. We can do this by using the velocity equation to find an equation for position, and find when that position reaches the ground (0 m ). To find the equation for position, we can integrate the equation for the velocity.

S = Position = Integrate(52.3 - 9.81t) = 52.3t - 4.8*t^2 + C

We need to find what C is to make this equation complete. We know that at t = 0, S = 1. Using this we can solve for C:

S = 1 = 52.3*0 - 4.8*0^2 + C

C = 1

So position becomes:

S = 52.3t - 4.8*t^2 + 1.

Now let's set position equal to 0, and solve for time:

S = 0 = 52.3t - 4.8t^2 + 1

t = -.02, 10.7 sec

We don't care about the -.02 seconds, because we don't use negative time. This means that the ball is in the air for a total of 10.7 seconds before hitting the ground. To find the horizontal distance traveled, we just multiply the time in the air, buy the horizontal velocity we found earlier:

d = v*t = 20.1 * 10.7 = 215.07 m

The ball travels 215.07 meters.