Projectile Motion: Throwing a Package to Reach a 2nd Story Window

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To determine the speed and angle required to throw a package to a second-floor window, the problem involves calculating the vertical and horizontal distances. The package must travel a vertical height of 2.7 meters (4.2m - 1.5m) and a horizontal distance of 3.0 meters. The vertical velocity will be zero at the maximum height, which can be used to find the initial vertical velocity. By understanding the relationship between the distances and the projectile motion equations, one can derive the necessary initial speed and launch angle. This approach emphasizes the importance of analyzing both vertical and horizontal components of the motion.
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Homework Statement


Standing on the ground 3.0m from the wall of a building you want to throw a package from your 1.5m shoulder level to someone in a second-floor window 4.2m above the ground. At what speed and angle should you throw the package so it just barely reaches the window?


Homework Equations


y=tan\vartheta-\frac{g}{2v^{2}cos(\vartheta)^{2}}x^{2}


The Attempt at a Solution


I'm not exactly sure where to begin with this problem. I tried drawing a picture of the problem but I'm still stuck. The only thing I know is that since the package just reaches the 2nd story window, the velocity Vy0=0 at some time t>0.
 
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Correct the vertical velocity will be zero at some time after t>0. If it just goes through the window this will be its maximum height.

Start by thinking about.

{1} what is the change height the ball travels through to get to the window?

{2} How far does the ball move horizontally?

Knowing the height the ball moves to before stopping allows you to calculate the initial vertical velocity, which then will allow you to work out the time.
 
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