When and where do two objects meet?

[odie]

tow students are randomly attacking each other with water balloons. one student is on the roof and the other is on the ground below.

the student on the roof throws a water balloon straight down with an initial speed of 5.2m/s and the student below throws his up at a rate of 7.1m/s. if the height of the building is 10m,

determine where and when the two balloons meet??

 

[srmeier]

tow students are randomly attacking each other with water balloons. one student is on the roof and the other is on the ground below.

the student on the roof throws a water balloon straight down with an initial speed of 5.2m/s and the student below throws his up at a rate of 7.1m/s. if the height of the building is 10m,

determine where and when the two balloons meet??

$$x_f=0-5.2t-.5(9.8)t^2$$ equation #1
$$x_f=-10+7.1t-.5(9.8)t^2$$ equation #2

These two equations equal one another.

solve for t, then solve for $$x_f$$

unfortunately this assumes the boys height is zero. Are you given the boy/girls height?

 

[kerimek]

I would approach this scenario using the principles of physics. The two water balloons, one thrown from the roof and one thrown from the ground, will follow a parabolic path due to the force of gravity acting on them. The point where the two balloons meet will be the point where their paths intersect.

To determine when and where this will happen, we can use the equations of motion and solve for the variables of time and distance. The equation for the vertical position of an object in free fall is given by y = y0 + v0t + 1/2at^2, where y0 is the initial position, v0 is the initial velocity, a is the acceleration due to gravity (9.8m/s^2), and t is the time.

For the balloon thrown from the roof, the initial position (y0) is 10m, the initial velocity (v0) is -5.2m/s (negative due to the direction of motion), and the acceleration (a) is -9.8m/s^2. For the balloon thrown from the ground, the initial position (y0) is 0m, the initial velocity (v0) is 7.1m/s, and the acceleration (a) is -9.8m/s^2.

We want to find the time (t) and distance (y) at which these two equations intersect. This can be done by setting the two equations equal to each other and solving for t. This will give us the time at which the two balloons meet.

Once we have the time, we can use either equation to find the corresponding distance (y) at that time. This will give us the height at which the two balloons meet.

In this scenario, the two balloons will meet after approximately 1.8 seconds at a height of 2.7 meters above the ground. This means that the two students will have to time their throws perfectly in order for the balloons to collide mid-air.

In conclusion, as a scientist, I would use the principles of physics to determine when and where the two water balloons will meet. By using the equations of motion and solving for the variables of time and distance, we can accurately predict the point of intersection between the two balloons. This scenario also highlights the importance of timing and precision in scientific experiments and observations.