Projectiles II

[Heat]

1. The problem statement, all variables and given/known data

A man stands on the roof of a building of height 14.1 m and throws a rock with a velocity of magnitude 31.6 m/s at an angle of 25.0 above the horizontal. You can ignore air resistance.

Calculate the maximum height above the roof reached by the rock.

Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

2. Relevant equations

V^2 = V initial ^2 + 2a(x-xo)


3. The attempt at a solution


The first part is the part, that I am getting stuck on, the other two parts of the question following it hopefully I will be able to do, as I encountered that type of problem before.

Anyways, my attempt:

Max height = unknown
Angle : 25 degrees
Vy = 0 m/s (at the highest point, Velocity is 0.
Vy initial is 13.35 m/s. ( I got this using 31.6 sin 25)
Height at beginning is 0. (Ignoring the fact that's the building is 14.1m)

0 = 13.35^2 + 2 (-9/8)(y-0)
-178.22 = -19.6y
y = 9.09 m

does this seem right?

[hage567]

In your initial statement of the question you say the angle is 25 degrees, but you use 75 in your calculation. Your approach works, though.

[Heat]

oops, my messy handwriting when working it out. does that seems right now..

[hage567]

Looks OK to me.

[Heat]

I thought this was going to be simple, but I was wrong:

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

my work:
V^2 = 13.35^2 + 2 ( -9.8) (9.10)
V^2 = -.1375:confused:

[learningphysics]

I thought this was going to be simple, but I was wrong:

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

my work:
V^2 = 13.35^2 + 2 ( -9.8) (9.10)


vertical displacement is -14.1m. not 9.10

remember that this is only vertical velocity... you'll need to get the total velocity after getting this part...

You can also use conservation of energy to do this problem.

[Heat]

ok I got Vy = 21.32.

i would think vx remains the same so

it would be

sqrt(21.32^2 + 13.35^2)
= 25.15
??

[learningphysics]

ok I got Vy = 21.32.



yes, but Vy = -21.32 (makes no difference for the magnitude)



i would think vx remains the same so

it would be

sqrt(21.32^2 + 13.35^2)
= 25.15
??

But vx is not 13.35. that is the initial vy.

[Heat]

you are right, my messy handwriting at it again.

Vx is 28.64.

so the total velocity should be: 35.70
which is right.

I will now try to solve:

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

hopefully this is the same as a question I asked before regarding horizontal distance. :)

[learningphysics]

you are right, my messy handwriting at it again.

Vx is 28.64.

so the total velocity should be: 35.70
which is right.

I will now try to solve:

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

hopefully this is the same as a question I asked before regarding horizontal distance. :)

yes. looks right. a faster way to solve the problem is with conservation of energy if you've covered energy...

[Heat]

I guess it will not work, as the previous was free falling with 0 initial velocity.

Here is what I did:

to find time it takes to drop I used the following:

-14.41 = .5(-9.8)t^2
t = 1.696

then I did

xfinal = 1.696/2 (28.64+28.64)
= 48.57

which is wrong :(

[learningphysics]

I guess it will not work, as the previous was free falling with 0 initial velocity.

Here is what I did:

to find time it takes to drop I used the following:

-14.41 = .5(-9.8)t^2
t = 1.696

then I did

xfinal = 1.696/2 (28.64+28.64)
= 48.57

which is wrong :(


I'm confused. Is this the same problem?

[Heat]

sorry, this is to find the horizontal distance from the base of the building to the point where the rock strikes the ground. In this same problem.

[learningphysics]

sorry, this is to find the horizontal distance from the base of the building to the point where the rock strikes the ground. In this same problem.

Use d = v1*t + (1/2)(-g)t^2

to get time.

[Heat]

ok I got time of 3.85s

using -14.1 = -31.6t -9.8t^2

now
it should be hopefully the following

xfinal = 3.85/2 (28.64+28.64)
= 110.16

seems awfully large.

[learningphysics]

ok I got time of 3.85s

using -14.1 = -31.6t -9.8t^2



No. it should be -14.1 = 31.6sin(25)t - 4.9t^2



now
it should be hopefully the following

xfinal = 3.85/2 (28.64+28.64)
= 110.16



I don't understand what you're doing here.... once you have the time... the horizontal distance is just 31.6cos(25)*t

[Heat]

ok for t i got 3.54s

now when I plug in the equation provided for horizontal distance, I get 31.6cos(25) *3.54 = 101.38

which is correct.

also

prior to my equation before...what I did was the following

xfinal = time/2 (vox+vx)

which I retried and gave me the same thing, the problem I found out is that I had time wrong. :redface:

thank you for your help.