A "lovesick" individual wants to throw a bag of candy and love notes into the open window of their significant other’s bedroom 10.0 m above. Assuming it just reaches the window, they throw the gift at 60.0o to the ground: At what velocity should they throw the bag? How far from the house are they standing when they throw the bag? ( A. 16.2m/s B. 11.5m)

Explanation:

Let the velocity be v .

vertical component of the velocity = v sin 60 = √3 v /2

it reaches maximum height of 10 m .

v² = 2 gh

( √3 v/2 )² = 2 x 9.8 x 10

3 v² = 196 x 4

v² = 65.33 x 4

v = 16.2  m /s

Let time taken to reach height of 10 m

v = u - gt

v sin 60 = 9.8 t

16.2 x √3  /2 = 9.8 t

t = 1.43 s

horizontal distance covered = v cos 60 x t

16.2 x .5 x 1.43 = 11 .5 m

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.