A bullet is fired from the ground at an angle 45 to the power of ring operator above the horizontal. What initial speed V subscript 0 must the bullet have in order to hit a point 125 meters high located 150 meters away? Use 10 space text m/s end text squared for gravity acceleration.

The initial speed of the bullet is 30 m/s

Explanation:

Given;

angle of projection, θ = 45°

Apply kinematic equation, to determine time to reach vertical height of 125 meters;

Y = Vt + ¹/₂gt²

At maximum height, v = 0

125 = ¹/₂ x 10t²

125 = 5t²

t² = 125/5

t² = 25

t = √25

t = 5 seconds

Finally, the magnitude of the initial speed of the bullet to reach 150 meters horizontal distance

X = V₀t + ¹/₂gt²

gravity has little or no influence in horizontal displacement

X = V₀t

150 = V₀ x 5

V₀ = 150/5

V₀ = 30 m/s

Therefore, the initial speed of the bullet is 30 m/s

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.