Aball is thrown upward at an unknown angle with an initial speed of 20.0 m/s from the edge of a 45.0-m-high cliff. at the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. the woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored. at what angle above the horizontal should the ball be thrown so that the runner catches it just before it hits the ground, and how far does the woman run before she catches the ball?

distance by the runner = 6 t

θ = 72.54°

Explanation:

initial velocity of ball = 20 m/s

speed of the runner = 6 m/s

horizontal velocity in the projectile = 20 cos θ

horizontal distance traveled by the ball =  20 cos θ × t

t is the time at which the velocity is to be calculated

distance by the runner = 6 × t

to catch the ball distance by the ball should be equal to the distance traveled by the runner.

20 cos θ × t = 6 × t

cos θ =  0.3

θ = 72.54°

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.

the answer would most likely be c, because if it's a comical piece of information, it may likely not be true, if it's a site that allows anybody to edit the content, it could mean a five year old is typing the info, and if it's a tv show produced by a celeb, it may have a biased opinion.