Projectile Motion and Circular Motion

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Kushal
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Homework Statement



I. A projectile is launched at an angle 'alpha' to the horizontal with an initial velocity u. Write down expressions for (1) horizontal velocity (2) total time of flight. Combine these expressions to show that R = (u^2 sin 2'alpha')/g.

II. Over a period of 5.0s a point changes its velocity from 10m/s at a bearing of 90 degrees to 10m/s at a bearing of 150 degrees.

Calculate:
(a) change of velocity
(b) the average acceleration


Homework Equations



I. Equatons of motions??!
II. v = rw
a = v/r^2

The Attempt at a Solution



I. I got the first one, Ux = u cos 'alpha'. I cannot find an appropriate equation for solving for time. This is blocking the whole problem for me.

II. change in velocity should be zero
 
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Kushal said:

Homework Statement



I. A projectile is launched at an angle 'alpha' to the horizontal with an initial velocity u. Write down expressions for (1) horizontal velocity (2) total time of flight. Combine these expressions to show that R = (u^2 sin 2'alpha')/g.

I. I got the first one, Ux = u cos 'alpha'. I cannot find an appropriate equation for solving for time. This is blocking the whole problem for me.

sin (2A) = 2sin(A)cos(A)
now try. :smile:
 
1.get the y component of speed .also calculate the y component of disp at any time t remember the motion along y is an accelarated one.equtae that to 0 for finally the y component is 0.
2.remeber velocity is a vector quantity.so change in velocity can also be there is the direction changes. and average accelaration is total change in velocity by total time
 
I. R is actually the range. So i should be concentrating on the x components. I need time in terms of the x components. I can use the constant speed formula to find t, but i also get the unknown range.

II. I know that there is a change in direction but not magnitude. Velocity changes and there is acceleration. What i want to know is how to express a change of velocity when there is only a change of direction but not magnitude?
 
Kushal said:
I. R is actually the range. So i should be concentrating on the x components. I need time in terms of the x components. I can use the constant speed formula to find t, but i also get the unknown range.

You have to use vertical displacement:
[tex]v_{vertical}*t + (1/2)(-g)*t^2 = 0[/tex]

Solve for t.

II. I know that there is a change in direction but not magnitude. Velocity changes and there is acceleration. What i want to know is how to express a change of velocity when there is only a change of direction but not magnitude?

Use vectors... draw the triangle for addition of vectors... and solve for the difference in the two vectors. Then divide by time for average acceleration.