Projectile Motion Problems: Calculating Maximum Angles and Heights

[hancyu]

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.

 

[Dick]

Yep, ur clueless. Pick one of these problems and TRY to do something with it. Anything. Then someone can help you.

 

[hancyu]

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...

 

[Dick]

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.

 

[hancyu]

**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?

 

[hancyu]

help on the first problem please. i really don't get the question.

 

[Dick]

...

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.

 

[hancyu]

sorry...i was trying to solve and did not notice that you posted a new msg.

 

[hancyu]

^^ i edited that part. i tried solving without the (b)a(/b)

 

[Dick]

**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.

 

[Dick]

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.

 

[hancyu]

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...

 

[hancyu]

problem #2

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

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

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

 

[Dick]

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.

 

[Dick]

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.

 

[hancyu]

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?

 

[hancyu]

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

 

[hancyu]

is this anything like.
y=yo+Vyot +1/2gt^2
x=xo+Vxot+1/2gt^2

then u get: (y-x)/(Vyo - Vxo) = t
...

 

[Dick]

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.

 

[hancyu]

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?

 

[Dick]

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?

 

[hancyu]

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?

 

[hancyu]

^^ i mean... T-mgsinO + mgcosO(.2) = 0

 

[Dick]

^^ 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.

 

[hancyu]

so...i equated both of them to each other. i got

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

then what?

 

[Dick]

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?

 

[hancyu]

there's tangent which is equal to sinO/cosO. is that it?

 

[Dick]

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.

 

[hancyu]

uhm...sin(90-O) = cosO or cos(90-O)= sinO

 

[Dick]

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).