Finding Magnitude of Velocity with Variables

[RedRumRiots]

Homework Statement

Romeo tosses a pebble at Juliet's window. Before crashing through the window the pebble is moving horizontally, having traveled a horizontal distance x and a vertical distance y. Find the magnitude and direction of the pebble's velocity as it leaves Romeo's hand.

Homework Equations

I know that Vx=VCosA and Vy=SinA.
Also, t=VSinA/g

The Attempt at a Solution

The attempts are pretty much any relevant equations I can find. There's no real math, except I know that the initial Vy is 0. I also figured if the projectile was graphed it would be the right side of the vertex of the parabola. This is why t=VSinA/g. I do not know where to go from here though. I was thinking of using y=VsinA*t-.5g(t)^2. but I really don't know

 

[Mentz114]

You know this -

time to hit window, t = x/Vx where Vx is the x velocity

Vy = gt, because Vy=0 at time t ( the sign may be wrong here, Vy is in the opposite direction to g)

y = 1/2 gt^2, distance traveled vertically in time t.

Some manipulation should give what you're looking for.

 

[highcoughdrop]

I don't understand this. How can we figure out what t is. I do understand that y = .5gt^2. Just not the rest of it. Any help here would be appreciated!

 

[alphysicist]

Hi RedRumriots,

Homework Statement

Romeo tosses a pebble at Juliet's window. Before crashing through the window the pebble is moving horizontally, having traveled a horizontal distance x and a vertical distance y. Find the magnitude and direction of the pebble's velocity as it leaves Romeo's hand.

Homework Equations

I know that Vx=VCosA and Vy=SinA.
Also, t=VSinA/g

The Attempt at a Solution

The attempts are pretty much any relevant equations I can find. There's no real math, except I know that the initial Vy is 0.

Why do you think the Vy=0 initially? You do know that Vy is zero right before the stone hits the window; what does that tell you about that point of the trajectory?

If Vy=0 initially, then this stone would never move upwards.