So I've been having problems with this physics 12 question because it wants me to find the initial velocity but it only gives dx, dy and the angle. How do I find time in order to figure out the initial velocity.

I don't have much fg to give but i'll give all of it for some help, much appreciated :))


http://imgur.com/9hEKZ9N

This post was edited by Jwoww on Apr 24 2014 11:44am

You can write two equations- one for y axis w/ gravity and one for x axis. Solve both in terms of time and solve for velocity

You don't need to actually find t, time

This post was edited by cialda on Apr 24 2014 12:50pm

Wouldn't I need 3 variables to solve for either x or y. What 3 variables would I need? For y I have a = -9.8, dy = -1.5m and what is the 3 variable I would use/how to obtain it.

no
in this stuff , things can be broken up into components
they do their own things
like, which hits the ground first
a horzontally fired bullet or a ball dropped from the same height
it is actually easier to do it in 1 d for this type of projectile problem

d = vt + 1/2at^2

i think you can just use the second half of it for the y component

-1.5m = 1/2 (-9.8)t^2

solve for t

then plug in to the first half of equation in terms of x

23 = vt

Not 100% sure though

So your solving for the y component... so is the initial velocity in the y equal to 0 even in an angled projectile launch? Because judging from the equation it would be 0?

This post was edited by Jwoww on Apr 27 2014 10:41pm

Totally forgot about this. I did the work but I'm not entirely sure if I'm doing this correctly. Can anyone verify my method/work? I put it on imgur

http://imgur.com/pPWO2jG

This post was edited by Jwoww on May 7 2014 02:33am

Let the cannon be the origin of a frame.
You're looking for the initial velocity V , that has 2 components :
Vx = V . cos(20)
Vy = V . sin(20)

According to Newton, and in the absence of any other force :

V(t)x = Vx
V(t)y = Vy - g.t

In other words, the horizontal velocity doesn't change all along the shot, because no horizontal force apply to the bullet.
And the vertical velocity is affected by gravity, with g ~ 9.8 m/s²

The position of the bullet can be expressed, by taking the anti-derivative of the velocity, and by figuring out the initial position (which is (0;0)) :

M(t)x = Vx.t
M(t)y = Vy.t - g.t²/2

And you know that, at a given moment, the projectile reaches the position ( 23 ; -1.5 ) :

M(t)x = 23
M(t)y = -1.5

You can solve for t the horizontal position :
Vx.t = 23
t = 23 / Vx
t = 23 / (V.cos 20)

And plug in the result into the expression of the vertical position at that moment :
Vy.23 / (V.cos 20) - g ( 23 / (V.cos 20) ) ² / 2 = -1.5

Notice that Vy = V.sin 20, and that sin/cos = tan :

g( 23 / (V.cos 20) ) ² / 2 = 23.tan 20 + 1.5

And you're now able to solve for V this equation :

V = (23/cos 20). sqrt ( g / (46.tan 20 + 3) )

Using 9.8 m/s² for g, you should find an initial velocity of V ~ 17.34 m/s.

This post was edited by feanur on May 7 2014 04:36am