Question

Picture attached.

Answer 1

@Mertsj @.Sam. @thomaster

Answer 2

@UnkleRhaukus

Answer 3

which bit are you up to

Answer 4

I do not know whether I've started correctly, honestly.

Answer 5

Can we do this a little bit faster?

Answer 6

what equations do you have

Answer 7

I need to create vector here, right?

Answer 8

I actually am not sure how to graph this.

Answer 9

Can you help me?

Answer 10

have you found h?

Answer 11

Not yet, actually. I just seen this problem right now. I'm just renting at computer shop.

Answer 12

consider the vertical and horizontal motion independently

Answer 13

Yes. But I am not sure with its graph.

Answer 14

Can you teach me? Don't worry I will not just copy solutions and answers. I will understand the concept you'll gonna teach.

Answer 15

Hence h = 38tan40

Answer 16

\[x(t)=x_0+{v_x}_0t+\tfrac12a_xt^2\]
\[y(t)=y_0+{v_y}_0t+\tfrac12a_yt^2\]

Answer 17

\[y(t) = 0+0+\frac{ 1 }{ 2 }(9.8)(2.15)^2 = 22.65m\]
Is this right?

Answer 18

@UnkleRhaukus

Answer 19

yep that is h

Answer 20

now look at the x- equation to find v_x_0

Answer 21

Oh yeah. I got this, I think.
\[\[V _{xo} = \frac{ 38 }{ 2.15 } = 17.6m/s\]]

Answer 22

yeap

Answer 23

Then I'll use
\[V _{fy} = v _{oy} + at = 0+(9.8)(2.15) = 10.54\]
\[v _{fx} = v _{ox} + at = 17.67 + (9.8)(2.15) = 38.74\]
Hence, getting the sqrt of the sum of those final velocities.
\[v _{f} = 40.15m/s\]
Am I right?

Answer 24

And the angle would be:
\[\alpha = \tan ^{-1}\frac{ 10.54 }{ 38.74 } = 15.22\] Is this right?

Answer 25

where is that angle?

Answer 26

the question at C.

Answer 27

hmm i guess my diagram was wrong then

Answer 28

but that would mean there was some \({v_y}_0\)

Answer 29

But still, our h is correct right? But how can we solve for Vyo?

Answer 30

I'm in a rush, and I really need to understand the concept and the solution here.