# Cannon projectile motion problem

1. Nov 28, 2004

### Jameson

A daredevil is shot out of a cannon at 24.0° to the horizontal with an initial speed of 26.0 m/s. A net is positioned a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the daredevil?

Ok, here's how the teacher explained it.

Vx = Vcos(theta)
Vy=Vsin(theta)

t = (horizontal distance) / Vx

and finally y = VyT + .5gt^2

------------
So, here's the work.

Vx = 23.75
Vy = 10.575

t = 2.105

y = (10.575)(2.105) + .5(-9.8)(2.105)^2

= .54797 m

Help me :surprised

2. Nov 28, 2004

### prasanna

Basically this is a projectile motion problem.

In a projectile motion, the motion is split up into 2 parts - Horizontal and vertical. That's what you teacher has done here (Vcos(theta) and Vsin(Theta)).

You got to remember that the two motions are independent of each other,
so we can apply equations of motion to both.

What we have to do in this problem is :
When the dare devil has travelled 50 m horizontally, we need to find his vertical position. This is where the net is to placed, right?

So, That's all!!

3. Nov 28, 2004

### Jameson

Yeah, I understand that the two motions are independent of each other. I just don't see how to relate them so I can get the correct answer. Can someone show me where I went wrong?

Thanks

4. Nov 29, 2004

### cdhotfire

Okay first you got that the Vi is 26 m/s, correct?
so we need to divide that into 2 components, vertical, horizontal.
so

---------/
--------/-| Vertical Vi, refered to as Vy sin 24*=26(hypo)/Vy
-------/--|
------/___|
-----24*---Horizontal Vi, refered to as Vx cos 24*=26(hypo)/Vx

so we got

H-|a=0---|Vix=23.75-|x=50
---------------------------------------
V-|a=-9.8|Viy=10.575|x=?

we have three things on the horizontal so we find time with x=Vit + (1/2)at^2

a=0 so we are left with 50(distance)=23.75(Vix)t
t=2.105 s

now we have 3 things on the bottom so we can solve for x with the same equation

x=10.575(2.105) - (1/2)(9.8)(2.105)^2
so we get x=.55

5. Nov 29, 2004

### Zlex

Doesn't, acceleration due to gravity opposes the motion?

Edit: n/m, you just moved the negative sign.

Last edited: Nov 29, 2004
6. Nov 29, 2004

### cdhotfire

hmm i might have confused him, thanks for clarifying that.