Projectile Motion Problems: Calculating Maximum Angles and Heights

In summary: If you have a mass on an incline, and you resolve the weight force into two components, what is the direction of the normal force?I would say perpendicular to the incline, and equal in magnitude to the component of the weight force perpendicular to the incline.This is the same as the force the incline exerts on the mass, no?Edit: This is a reply to the post below, and the other one. I am getting very confused by the way this forum works.i'm sorry...i'm just really bad at physics...It's ok. I'm sorry too. I'm getting confused by the way this forum works. I would like to help you with your physics but
  • #1
hancyu
57
0

Homework Statement



1. Find the maximum angle of projection of a projectile such that its position vector from the origin to the subsequent position of the projectile is always increasing.

2. Consider two masses at either end of a frictionless pulley. The first block of mass 10kg sits on a inclined plane while the other block of mass 5kg hangs on the other end of the pulley. The coefficient of kinetic friction between the plane and the block is 0.2. Find the angle of inclination of the plane if the masses are to move with constant velocity.

3. Find the angle of projection of a projectile if its horizontal range will be equal to the magnitude of the maximum height reached.

4. A projectile is launched at an angle of (degrees)40 with an initial velocity of 25m/s when upon traveling 22m horizontally it strikes a wall. Find the height and velocity of the projectile when it hits the wall.

Homework Equations


i don't know how to do all these...

The Attempt at a Solution


i'm clueless.
 
Last edited:
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  • #2
Yep, ur clueless. Pick one of these problems and TRY to do something with it. Anything. Then someone can help you.
 
  • #3
im trying to solve number4

this is what i got...

Vxo= 25cos40 = 19.15m/s
Vyo= 25sin40 = 16.07m/s

tx=ty= (Vyo)/(9.8) = 1.64s

thus, a = (Vxo)/(1.64) = 11.68m/s^2

using x=Vxot +1/2at^2...
x=47.11

im lost...
 
  • #4
Good! You got the initial velocities ok. From there you don't derive the acceleration. The acceleration is given, it's just gravity -9.8m/sec in the y direction. So there is no acceleration in the x direction:

x(t)=v0x*t

And in the y direction:

y(t)=v0y*t+(1/2)*g*t^2

Use one of those equations to find the time it hits the wall. Which one? It's 22m HORIZONTALLY away.
 
  • #5
**EDIT**

so... 22/19.15 = t = 1.15s

y=Vyot +1/2gt^2
y=12.0m

vyf^2 - vyo^2 = 2gy

vyf=4.8m/s

is this right?
 
Last edited:
  • #6
help on the first problem please. i really don't get the question.
 
  • #7
hancyu said:
...

using x=Vxot +1/2at^2...
x=47.11
y=13.18

22m/(47.11/2) = 93%

y*93% = 12.26m = h

2(12.26)(-9.8) - 16.07^2 = -Vyf^2

Vyf = 22.32 m/s

is this right?

That's AWFUL. It doesn't make any sense at all. Why did you ignore what I wrote? I thought you wanted help.
 
  • #8
sorry...i was trying to solve and did not notice that you posted a new msg.
 
  • #9
^^ i edited that part. i tried solving without the (b)a(/b)
 
  • #10
hancyu said:
**EDIT**

so... 22/19.15 = t = 1.15s

y=Vyot +1/2gt^2
y=12.0m

vyf^2 - vyo^2 = 2gy

vyf=4.8m/s

is this right?

That's it. Good.
 
  • #11
hancyu said:
help on the first problem please. i really don't get the question.

If you throw a ball upwards at a steep angle it will go away from you and then come back towards you (and hit you on the head if the angle is too steep). If you throw it out at a shallow angle then it always travels away from you and doesn't come back. They want you to find the angle that divides the two cases.
 
  • #12
Dick said:
If you throw a ball upwards at a steep angle it will go away from you and then come back towards you (and hit you on the head if the angle is too steep). If you throw it out at a shallow angle then it always travels away from you and doesn't come back. They want you to find the angle that divides the two cases.

okayyy... so if angle=0 range =0...
angle=45 range = maximum...

then...
 
  • #13
problem #2

g*[(5-10cosO)/(5+10)] = 0

a=0 because its' velocity is constant. ryt?

then... angle should be...60degrees?
 
  • #14
hancyu said:
okayyy... so if angle=0 range =0...
angle=45 range = maximum...

then...

No, it's not 45 degrees. If the launch angle is theta then you work out x(t) and y(t). The distance squared from the launch point (0,0) is r(t)^2=x(t)^2+y(t)^2. You want to take the derivative of r(t)^2 with respect to t and then figure out for which angles theta that expression has a zero derivative. These are the ones where it comes back towards you.
 
  • #15
hancyu said:
problem #2

g*[(5-10cosO)/(5+10)] = 0

a=0 because its' velocity is constant. ryt?

then... angle should be...60degrees?

Yes, the forces balance because a=0. You might want to write out explicitly what the forces acting on the mass on the incline are. You've left out the kinetic friction.
 
  • #16
Dick said:
Yes, the forces balance because a=0. You might want to write out explicitly what the forces acting on the mass on the incline are. You've left out the kinetic friction.

so...
T-m = ma
T-10cosOg + 10cosO(.2) = -ma

mg+mcosOg +10cos(.2)=(ma +ma)

then... g(m +mcosO) + 10cosO(.2)=a(m+m)

how will i get O?
 
  • #17
Dick said:
No, it's not 45 degrees. If the launch angle is theta then you work out x(t) and y(t). The distance squared from the launch point (0,0) is r(t)^2=x(t)^2+y(t)^2. You want to take the derivative of r(t)^2 with respect to t and then figure out for which angles theta that expression has a zero derivative. These are the ones where it comes back towards you.

i really don't get this.sorry:confused:
 
  • #18
is this anything like.
y=yo+Vyot +1/2gt^2
x=xo+Vxot+1/2gt^2

then u get: (y-x)/(Vyo - Vxo) = t
...
 
  • #19
hancyu said:
so...
T-m = ma
T-10cosOg + 10cosO(.2) = -ma

mg+mcosOg +10cos(.2)=(ma +ma)

then... g(m +mcosO) + 10cosO(.2)=a(m+m)

how will i get O?

I don't understand how you are getting these funny equations. Could you explain in addition to just writing a bunch of stuff down? The mg force down the incline should be related to sin(O). The frictional force should be related to the normal force which involves cos(O). You should have both in the equation.
 
  • #20
the forces acting on mass=5kg is...tension and gravity ryt...
so, T- mg= ma
then, the forces acting on mass=10kg is tension, gravitational force,frictional force
so, T-mgcosOg - ________ + mcosO(.2)= -ma

how do i get normal force?
 
  • #21
hancyu said:
the forces acting on mass=5kg is...tension and gravity ryt...
so, T- mg= ma
then, the forces acting on mass=10kg is tension, gravitational force,frictional force
so, T-mgcosOg - ________ + mcosO(.2)= -ma

how do i get normal force?

Ok, you've got the forces names right. T is 5kg*g. O is the angle the incline makes with the horizontal, yes? I would say the component of the weight down the incline is mgsin(O), and yes, the frictional force is mgcos(O)*(.2). That just comes from splitting the weight force up into parts tangential to the incline and normal to it. a=0. You keep writing down stuff that is halfway correct, but I get the feeling you really don't understand it, because of stuff that might just by typos. Like "mgcosOg" has two g's in it. "mcosO(.2)" doesn't have any. How can that be, they are both dimensionally wrong. One of the trig functions is wrong. And there is nothing to put in the blank space. Are you sure you shouldn't be warming up on some easier questions?
 
  • #22
these questions are for extra credit(which i badly need.so, they shoudnt be as easy as cake.)

**
so, T-mg=0 and T-mgcosO +mgcosO(.2)=0

is that right?
 
  • #23
^^ i mean... T-mgsinO + mgcosO(.2) = 0
 
  • #24
hancyu said:
^^ i mean... T-mgsinO + mgcosO(.2) = 0

Extra credit, that explains it. Does that seem right to you? Is the magnitude of every force correct? Do they point in the right direction? .... Yes, it's right.
 
  • #25
so...i equated both of them to each other. i got

-5g = -10gsinO + 10(.2)gcosO

then what?
 
  • #26
First cancel g. Now all that you don't know are sin(O) and cos(O). Is there a relation between sin and cos you can use to get rid of one of them?
 
  • #27
there's tangent which is equal to sinO/cosO. is that it?
 
  • #28
hancyu said:
there's tangent which is equal to sinO/cosO. is that it?

That would work if the -5 weren't there. No, I want a RELATION between sin and cos. They aren't independent.
 
  • #29
uhm...sin(90-O) = cosO or cos(90-O)= sinO
 
  • #30
hancyu said:
uhm...sin(90-O) = cosO or cos(90-O)= sinO

True. But that doesn't make it any easier to solve. How about sin(O)^2+cos(O)^2=1. Or sin(O)=sqrt(1-cos(O)^2).
 
  • #31
i don't get it... so, should i substitute sinO with (1-cosO^2)^1/2 ?
 
  • #32
i think theta should be 40.7(degrees)

i got it through trial and error. i just don't know how to solve for it.
 
  • #33
Good job on getting the approximation. Write the equation as 10*sin(O)-5=2*cos(O). Now put cos(O)=sqrt(1-sin(O)^2). Now square both sides. You get a quadratic in sin(O). Solve for sin(O). Since you squared both sides check that each of your answers actually satisfy the original equation. Now find O.
 
  • #34
so 1st i divided the whole equation with -5
i got.
1=2sinO -cosO then, i substituted sqrt(1-sin(O)^2) to cosO
then i got. 1=4sin^2(O) + 1 - sin^2(O) which doesn't make sense.

how should i do it?
 
  • #35
hancyu said:
so 1st i divided the whole equation with -5
i got.
1=2sinO -cosO then, i substituted sqrt(1-sin(O)^2) to cosO
then i got. 1=4sin^2(O) + 1 - sin^2(O) which doesn't make sense.

how should i do it?

You aren't doing the algebra right. If you square 2sinO-cosO you get 4sin^2(O)+cos^2(O)-4*sin(O)*cos(O). There's a cross term you've forgotten and you haven't gotten rid of the square root at all. Get the sqrt part all by itself before you square.
 

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