Question

a block of mass 5 kilograms moves to the right with constant velocity 8m/s and falls off a cliff of height 200m. 1) How long will it take to hit the ground? 2) What is the distance from the bottom of the cliff where it will land? 3) How fast will the block be moving just before it hits the ground?

Answer 1

ok are you familiar with projectile motion, kinematics, and vector components?

Answer 2

yeah I am

Answer 3

ok, so you're familiar with x and y directional motion being independent of one another right?

Answer 4

uh I only know the general formulas I'm not sure what you're talking about

Answer 5

ok so right now, while the block is on the cliff, it has no motion in the y-direction, only x-direction.

This means that when it shoots off the cliff, it's initial velocity in the y-direction is 0 even if it's initial velocity in the x-direction is 8m/s

You have to calculate the motion in the x and y directions separately. Make sense?

Answer 6

oh alrite yeah

Answer 7

so right now, the easiest thing to do is find the time it takes to hit the ground first. For this use the kinematic equations and do the calculation in the y-direction. I recommend using this one:
\[\huge y_f=y_i+v_yt+{\frac 1 2}a_yt^2\]

Answer 8

know what to substitute?

Answer 9

200= -4.9t^2 ?

Answer 10

would yf be -200?

Answer 11

close. your signs are off. time is never negative; besides, you can't take the square root of a negative number

your final position is 0, your initial position is 200, and your gravitational acceleration is -9.8m/s^2

Answer 12

make senese?

Answer 13

sense*

Answer 14

oh so initially its starting from 0 height to final being 200?

Answer 15

no no,

Answer 16

\[\large y_f=0m\]
\[\large y_i=200m\]
\[\large g=-9.8m/s^2\]
\[\large {v_y}_i=0m/s\]

Answer 17

do these numbers make sense?

Answer 18

yes! they do thank you

Answer 19

i got 6.3s for time

Answer 20

Don't look at me; that is what I got but is it what's in your solution (if you have it)?

Answer 21

yeah after taking the squareroot of 400/ 9.8

Answer 22

I meant the answer your book gives, if your book gives the answers I mean.

Answer 23

oh yeah the solution manual has 6.39 as the answer. But I was just confused about taking x and y independently but you cleared that up :)

Answer 24

ok cool so that is correct. Do you need help with (B) and (C)?

Answer 25

I think i got c but could you help me with b please

Answer 26

ok so part b is looking for, according to your picture, the length d

Answer 27

Now the time it takes to hit the ground has to remain consistent. That means that you already know how long it takes for the block to reach the distance d

Answer 28

right

Answer 29

if the block is moving at a constant velocity, what is it's acceleration?

Answer 30

there's no acc

Answer 31

exactly

Answer 32

\[\large x_f=x_i+v_0t+ {\frac 1 2}at^2\]

Answer 33

x= 0 +( 8 )(6.39) +0 making x final 51.12m

Answer 34

gotcha thank you! :)

Answer 35

That's what I say, what does the book say?

Answer 36

it says the same answer

Answer 37

cool.

Answer 38

so did you solve part c?

Answer 39

to be honest I got 62.61m/s but am not as sure on that as the other two

Answer 40

for c the answer is rounded to 63

Answer 41

oh well.
Did you need me to explain that one or have you got it?

Answer 42

uh could you explain it please I don't think i'm doing it right

Answer 43

ok for this one you want to work in the y-direction

Answer 44

\[\huge {v_y}_f = {v_y}_i + a_y t\]

Answer 45

In the y direction, the block is in free fall. So you don't have an initial velocity. Well you do but it's 0.

Answer 46

And acceleration is simply gravitational acceleration. You found time in part a

Answer 47

Make sense?

Answer 48

yeah but my professor used pythagorean theorem or something of that sort to find v

Answer 49

OH ok, I see; I messed up

Answer 50

Do you know how to find the magnitude of a vector?

Answer 51

a^2+b^2 =c^2 ?

Answer 52

not quite

Answer 53

ok so when you have a vector \[\huge \vec v\]
In order to break it down into it's two components
you have
\[\large v_x=\vec vcos\theta\]
\[\large v_y=\vec vsin\theta\]
What we've been doing in regards to velocity is using v_x and v_y.
Agreed?

Answer 54

yeah

Answer 55

now to get the final velocity you need to add the two components together, so while you are correct that it is the pythagorean theorem, you are looking for c.