SE111 homework assigned Week 5 block 5 by Dr. Rickert. Due Monday, October 11, 1999 by 5PM.

**Problem 1**

Determine the projection of the vector 4
**i**
+ 3
**j**
onto the vector -6
**i**
+ 3
**j**
.

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**restart;**

`> `
**with(linalg):**

Warning, new definition for norm

Warning, new definition for trace

`> `
**A := vector([4,3]): B := vector([-6,3]):**

`> `
**Am := sqrt(dotprod(A,A)); Bm := sqrt(dotprod(B,B));**

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**projab := evalm((dotprod(B,A)/Bm)*(B/Bm));**

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**Problem 2**

A mass of 10kg is placed on an incline of /7. The amount of force required to keep the object from sliding down the incline is the size of the projection of the force vector (weight of the object pointing in the direction of gravitational attraction) onto the incline (find a vector parallel to this direction). Determine how much force is required to keep the mass from sliding down the incline.

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**restart:**

`> `
**with(linalg):**

Warning, new definition for norm

Warning, new definition for trace

`> `
**fvec := vector([0,98]);**

`> `
**pvec := vector([(30*cos(Pi/7)),(30*sin(Pi/7))]);**

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**fvecM := sqrt(dotprod(fvec,fvec));**

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**pvecM := sqrt(dotprod(pvec,pvec));**

`> `
**projection := evalm((dotprod(pvec,fvec)/pvecM)*(pvec/pvecM));**

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**size := evalf(evalm(sqrt(dotprod(projection,projection))));**

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**Problem 3**

A ball is hit (still in a vacuum) at an initial height of 1 meter, with an initial speed of 40 meter per second and angle of inclination of /6. Assume that the only force acting on the ball is that of gravity (downward at 9.81 m/s/s). The fence is 440 feet away and is 5 meters high. Determine whether the ball clears the wall.

`> `
**sty:=-4.9*t^2+(40*cos(Pi/6))*t+1;**

`> `

`> `
**stx:=t->(40*sin(Pi/6))*t;**

`> `
**sol1 := solve(sty=4,t);**

The first time is on the way up and out, we want to analyze the ball as it comes down towards the wall, so ignore the first solution.

`> `
**stx(sol1[2]);**

`> `
**evalf((((440*12)*2.2)/100));**

When the ball is at a height of 5 meters, the ball has traveled 140 meters, the wall is only 116 meters away. So, yes, it clears the wall.

**Problem 4**

A ball is hit at a height of 1.8 meters with an intial velocity of 29.43 meters per second and angle of elevation of
/3. The ball is hit towards an outfielder standing 88.29 meters from the point of impact. The outfielder eyes are 1.8 meters above the ground.

Relative to the outfielder's eyes, the ball is an some angle of elevation
, where the angle
is a function of the time since the impact. Determine the function
in simplest form.

(hint: what is ?)

`> `
**restart:**

`> `
**with(linalg):**

Warning, new definition for norm

Warning, new definition for trace

`> `
**xt:=29.43*cos(Pi/3)*t;**

`> `
**yt:=-4.905*t^2+29.43*sin(Pi/3)*t;**

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**theta := arctan(yt/(88.29-xt));**

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**Problem 5**

A planet orbits a star in a circular orbit with a period of 1 year at a distance of 130 million kilometers.

Write a parametric (vector) representation describing the planet's motion relative to the fixed star.

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**restart:**

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**with(linalg):**

Warning, new definition for norm

Warning, new definition for trace

`> `
**pxt:=(130*cos(2*Pi*t));**

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**pyt:=(130*sin(2*Pi*t));**

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**plot([pxt,pyt,t=0..1]);**

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**Problem 6**

A moon orbits the planet of Problem 5 with a period of 1/27.2 years at a distance of 10 million kilometers. Write a parametric (vector) representation describing the moon's motion relative to the fixed star.

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**mxt:=10*cos((2*Pi)*27.2*t);**

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**myt:=10*sin((2*Pi)*27.2*t);**

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**plot([mxt,myt,t=0..1/27.2]);**

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**plot([(mxt+pxt),(myt+pyt),t=0..1],scaling=constrained);**

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