# Projectile motion cannon

1. Dec 4, 2016

### gungo

1. The problem statement, all variables and given/known data
The Great Projecto is fired from a cannon at a velocity of 29.8 m/s at an angle of 50.4 degrees. He lands in a net which is 2.0 m above his starting elevation. What is his velocity when he hits the net?

2. Relevant equations
sin=opp/hyp
cos=adj/hyp
3. The attempt at a solution
Using trigonometry I found that:
vy1: 22.96 m/s
vx1: 18.995 m/s
vx2: 18.995 m/s
And I know that
ay(acceleration of y)= -9.8m/s^2
Since the question tell us that the net is 2.0 m above the starting elevation, dy+2= the height of the net...but how will I find the starting elevation? I don't know where to start.

2. Dec 4, 2016

### kuruman

You cannot solve this problem using just trigonometry. You have to use the kinematics equations for projectile motion.

3. Dec 4, 2016

### gungo

But none of the equations let me find the displacement without having another piece of information?
v2^2=v1^2+2ad>>>>I don't have d
v2=v1+at>>>>>I don't have t
d=v1^2+1/2a(t)^2>>>>I don't have d or t
d=v2^2-1/2a(t)^2>>>>I don't have d t or v2
d=(v1+v2/2)t>>>>I dont have d or t

4. Dec 4, 2016

### PeroK

@gungo what about using the velocity components in the two directions?

5. Dec 4, 2016

### kuruman

You know the displacement and the initial velocity in the vertical direction. Use them to find the time of flight first. Having the time of flight, you cn find anything else you need.

6. Dec 4, 2016

### PeroK

You don't need the time of flight.

7. Dec 4, 2016

### kuruman

That's right, you don't. You still need a kinematic equation in the vertical direction.

8. Dec 4, 2016

### gungo

Got it, thanks!

9. Dec 5, 2016

### PeroK

Note that $v^2 =u^2-2gh$ works for projectile motion, where $h$ is the upwards displacement in the $y$ direction.

You can derive this by:

$v^2 = v_x^2 + v_y^2 = u_x^2 + u_y^2 -2gh = u^2 -2gh$

Using the constant velocity in the x direction and the acceleration of gravity in the y direction.

Note also that you didn't need the angle of projection. When you come to learn about energy you'll see why.

10. Dec 5, 2016

### kuruman

The problem is asking for the velocity, which means two components. Although you don't need the projection angle to find the final speed as @PeroK pointed out, you still need the angle to find how that speed is divided between the x and y directions.

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