help with projectile motion?

Question

I really dont get projectile motion, here is a sample problem, can u guys help me (seriously)

An airplane is diving at a speed of 83.3 m/s, and the angle between the plane's dive path and the vertical is 50 degrees. the plane launches a rocket which lands at 750m horizontally from beneath the launch point. (a) find the flight time of teh rocket (b) how high was the release point?


PLEASE HELP ME! I'M REALLY STUCK!

This is all the info i was given, in my other problem

Answer 1

Start off by drawing a right triangle. The hypotenuse of the triangle is the plane's/rocket's flight path straight into the ground (we'll assume the pilot pulls up after launching his rocket, of course). The angle is 50 degrees between the flight path and the vertical, so the remaining 40 degrees is the internal angle of the triangle representing the rocket's angle with the ground. In a right triangle, sine of the angle equals the ratio between the opposite side (the altitude of launch) and the hypoteneuse, and cosine of the angle equals the ratio between the adjacent side (the ground) and the hypoteneuse. Since you have a speed of 83.3 m/s, you can figure out the vertical and horizontal components of the plane's speed (falling at 53.54 m/s while travelling forward at 63.8 m/s, respectively). Assuming the rocket followed the exact path the plane would have taken if it crashed into the ground and that it travelled at the same speed of 83.3 m/s, you can use this to figure out what the rocket did -- you know it travelled 750 meters along the horizontal, so divide that by 63.8 m/s and you get 9 seconds of flight time. You know that if the plane had continued on it's course, therefore, it would have hit the ground in 9 seconds, so it took 9 seconds at 53.54 m/s to go from launch altitude to reach the ground, which means it started at 482 m.

Answer 2

First I would draw a picture of the plane, and draw a free bodyu diagram. The plane is at an angle from the vertical, so break up its velocity into two components - the vertical and horizontal. You will get: (rounding to 3 significant digits)

(total velocity) V = 83.3m/s
(vertical velocity) Vv = V*cos(50) ~ 53.5m/s
(horizontal vel.) Vh = V*sin(50) ~ 63.8m/s

From here, you just need to know the equation

x = Vo*t + (1/2)a*t^2 where x=distance traveled
vo=initial velocity
a=acceleration
t = time

and use it for both the horizontal and vertical directions.

horizontal: xh = Vh*t + (1/2)ah*t^2 (1)
vertical: xv = Vv*t + (1/2)av*t^2 (2)

You are asked for: flight time of the rocket - t
vertical height at launch - xv
Notice that "t" appears in both horizontal and vertical equations.

First, the horizontal:

You are also given that the rocket lands 750m horizontally from launch point, so horizontal distance xh = 750m. You also know that there is no acceleration in the horizontal direction, so ah=0. We calculated Vh before. So equation (1) can be solved for t:

750m = 63.8(m/s)*t --> t = 750/63.8 --> t~11.8

Now the Vertical:

Here, you have to know that the acceleration due to gravity is 9.8m/s^2. We have solved for Vv and t, so now we can solve (2):

xv = Vv*t+(1/2)*(9.8)*t^2
= 53.5*11.8 + (1/2)*(9.8)*(11.8)^2
xv ~ 1305m

(a) Asked for the flight time of the rocket: t = 11.8.
(b) Asked for height of release point: xv = 1313m.

Answer 3

The only way to solve this problem is to assume the rocket is launched at the same speed as the plane. If this is true, there is more than enough information to solve the problem:

a. t (total)= R/(v(o)cos50)

b. R = (v(o)cos50/g)[(v(o)sin50 + sqrt(2gz + v(o)v(o)sin50sin50)] Note: Solve for z in this equation. The last 2 terms in the equation, v(o) & sin50, are squared so I had to show them the adjacent to each other.

z-height of release point (initial height)
R- total range (750m)
v(o) - initial velocity (83.3 m/s)
g- gravity

Sorry if the formulas did not come out clear, but it is hard to show without some sort of equation editing software like in Word.

Answer 4

The unfavourable sign refers back to the process the strain, that's downward. by skill of convention, unfavourable velocity and unfavourable acceleration are downward. in case you placed g as effective, then a projectile released upwards might have unfavourable velocity, which does no longer make intuitive experience.

Answer 5

Not enough info, no wonder you're stuck.