# HW on projectile motion

1. Jul 22, 2008

### hancyu

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

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.

2. Relevant equations
i dont know how to do all these...

3. The attempt at a solution
i'm clueless.

Last edited: Jul 22, 2008
2. Jul 22, 2008

### Dick

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

3. Jul 23, 2008

### 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...

4. Jul 23, 2008

### 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.

5. Jul 23, 2008

### 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?

Last edited: Jul 23, 2008
6. Jul 23, 2008

### hancyu

help on the first problem please. i really dont get the question.

7. Jul 23, 2008

### Dick

That's AWFUL. It doesn't make any sense at all. Why did you ignore what I wrote? I thought you wanted help.

8. Jul 23, 2008

### hancyu

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

9. Jul 23, 2008

### hancyu

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

10. Jul 23, 2008

### Dick

That's it. Good.

11. Jul 23, 2008

### Dick

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. Jul 23, 2008

### hancyu

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

then...

13. Jul 23, 2008

### hancyu

problem #2

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

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

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

14. Jul 23, 2008

### Dick

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. Jul 23, 2008

### Dick

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. Jul 23, 2008

### hancyu

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. Jul 23, 2008

### hancyu

i really dont get this.sorry

18. Jul 23, 2008

### 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
...

19. Jul 23, 2008

### Dick

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. Jul 23, 2008

### 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?