A projectile is launched with an initial velocity of 25 m/s at an angle of 30° above the horizontal. The projectile reaches maximum height at point p then falls to point a, which is 65 m below the cliff. What is the horizontal distance that it travels (labeled x in the diagram)

The horizontal distance travelled by the object will be "111.07 m".

According to the question,

As we know,

→  

hence,

The horizontal distance will be:

→

or,

→

Thus the response above is right.  

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x ≈ 56 m

Explanation:

vertical initial velocity = = 25 m/s* sin(30°)= 12.5 m/s

height = h

t- time is found solving quadratic equation.

horizontal velocity =

Horizontal velocity is constant, so distance

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.