Find time of freefall based on little knowledge

[Planefreak]

Homework Statement

You fire a squirt gun horizontally from an open window in a multistory building and make note of where the spray hits the ground. Then you walk up to a window 5.0 m higher and fire the squirt gun again, discovering that the water goes 1.5 times as far. Ignore air resistance. How long does the second shot take to hit the ground?

Homework Equations

(delta)x = Vxt (delta)y = v_yit + 1/2gt²

The Attempt at a Solution

I've put some time into this problem. I am having trouble including the 5m into the rest of the equation. I understand that I need to write a system of equations but so far they have been unsuccessful. I've also tried assigning values to see how far the water will go then work from there.

I keep coming up with systems that equal zero. I am trying to include the 5 meters but that doesn't seem to help. Can someone point me in the right direction?

 

[mgb_phys]

The only tings you need to know.
Vertical acceleration is just 'g', horizontal acceleration is zero.
And you can handle vert/horiz motion separately.

The set of equations is then:
v = u + a t
s = ut+ 1.2 a t ²
v² = u² + 2 a s

Where u and v are initial and final velocity, s is distance, a is accelration

 

[thomate1]

To find the time of freefall in this scenario, we can use the equations for projectile motion. First, we need to establish a coordinate system and choose a direction as positive. Let's say the ground is our reference point and the direction of the squirt gun's motion is positive.

Using the first equation, (delta)x = Vxt, we can set up a system of equations by equating the distances traveled by the squirt gun in the two scenarios. We know that in the first scenario, the distance traveled is the same as the height of the building, which we can call h. In the second scenario, the distance traveled is 1.5 times the height of the building, or 1.5h.

So our first equation is: h = Vx1 * t1

And our second equation is: 1.5h = Vx2 * t2

We can also use the second equation, (delta)y = vyit + 1/2gt2, to solve for t2. In this case, the initial velocity in the y-direction is 0 since the squirt gun is fired horizontally. The acceleration due to gravity, g, is a constant -9.8 m/s^2. We can rearrange the equation to solve for t2:

t2 = (2*1.5h)/g

Now we can substitute this value of t2 into our first equation:

h = Vx1 * t1

h = Vx1 * [(2*1.5h)/g]

Solving for t1, we get:

t1 = (g*h)/(2*1.5h*Vx1)

Since we know the height of the building, h, and the velocity of the squirt gun in the first scenario, Vx1, we can plug in these values to find t1.

t1 = (9.8*5)/(2*1.5*Vx1)

Now we just need to solve for Vx1. We can use the fact that the distance traveled in the first scenario is equal to the height of the building, h. So we can set up another equation:

h = Vx1 * t1

Solving for Vx1, we get:

Vx1 = h/t1

Now we can plug this value into our previous equation for t1:

t1