Question

Read question below.

Answer 1

Find x in terms of h

Answer 2

and h2

Answer 3

@eashmore I evaluated it to -(h1-h2) BUT THAT MAKES NO SENSE

Answer 4

I used conservation of energy mgh=mgh2+1/2mv^2 and v=v0+at and dx=v0t+1/2at^2 and I got that answer. :( BUT ITS WRONG!

Answer 5

You're looking for the distance the box travels, x?

Answer 6

oh yeah

Answer 7

Your equations looks correct. Remember that the velocity has both a vertical and horizontal component.

We can find the time of flight from \[v_y = v \sin(45) - g t\]Since \(v_y=0\) at the top of the trajectory, \[t = {v \sin(45) \over g}\]It takes the same amount of time for the object to then reach the ground, therefore\[t = {2 v \sin(45) \over g}\]
Then the horizontal distance the box travels can be found as\[d_x = v \cos(45) t = v \cos(45) \left (2 v \sin(45) \over g \right) = {2 v^2 \over g} \cos(45) \sin(45)\]

Answer 8

By "v" you mean v init?

Answer 9

WHERED THE 2 COME FROM?

Answer 10

OH nevermind I gets it

Answer 11

v is the velocity of the box right before it takes off from the ramp.