These homework problems are due by 5PM, Friday, September 17, 1999.
**
**
They were assigned Week 2 Block 2 by Dr. Rickert.

Please print your Maple worksheet 2-up and highlight your answers.

**Problem 1**

A bowling ball is droppe
d from a height of 50 meters. Describe the height of the bowling ball as a function of time.

For what times is this model valid?

`> `
**restart;**

`> `
**g := -9.8:
so := 50:
vo := 0:**

`> `
**st := .5*g*t^2+vo*t+so;**

`> `
**plot(st,t=0..4);**

The times are only valid from t = 0 to t = 3.2

**Problem 2**

A rock is thrown from a height of 1 meters with an intial velocity of 20 meters per second upwards.

Describe the height of the rock as a function of time. For what times is this model valid?

With what velocity does the rock land?

`> `
**restart:**

`> `
**so := 1:
vo := 20:
g := -9.8:**

`> `
**st := .5*g*t^2+vo*t+so;**

`> `
**plot(st,t=0..5);**

`> `
**vt := 2*.5*g*t+vo;**

`> `
**land := st=0:**

`> `
**landt := solve(land,t);**

`> `
**landV := subs(t=landt[2],vt);**

The model is valid from times t = 0 to t = 4.1, when the ball lands at t = 4.1, the velocity of the rock is 20.5 m/s towards the earth.

**Problem 3**

An object is slid along a perfectly frictionless surface, so that it experiences no acceleration. It is given an initial velocity of 30 meters per second. How far is the objedct from it's starting point after 30 seconds?

`> `
**dist := 30*t;**

`> `
**subs(t=30,dist);**

After 30 seconds, the object is 900 meters from where it started.

`> `

**Problem 4**

A baseball is thrown so that the
*x*
-coordinate of it's position satifies the position function of Problem 3,

and it's
*y*
-coordinate satisfies the position function of Problem 2. Plot
*x*
(
*t*
) vs.
*y*
(
*t*
).

How far away does the ball land?

`> `
**subs(t=landt[2],dist);**

`> `
**plot([dist,st,t=0..5]);**

If the horizontal velocity of the ball (x-coord) was equal to the velocity of the object in Problem 3, and the vertical position of the ball (y-coord) was equal to the equation from Problem 2, then the ball would land 124 meters away.

**Problem 5**

A baseball is thrown from the point (0,0) with an initial speed of 40 meters per second at an initial angle of elevation of
.

Determine the baseball's position function as a function of time and
.

Plot the baseball's position function for at least two values of
on the same graph.

`> `
**restart:**

`> `
**xv := 40*cos(theta):**

`> `
**yv := 40*sin(theta):**

`> `
**syt := .5*g*t^2+voy*t+soy;**

`> `
**sxt := xv*t:**

`> `
**g := -9.8:
voy := yv:
soy := 0:**

`> `
**syt;**

`> `
**theta := Pi/6:
with(plots):
ang30 := plot([sxt,syt,t=0..6]):**

`> `
**theta := Pi/4:
ang45 := plot([sxt,syt,t=0..6]):**

`> `
**display({ang30,ang45});**

`> `

`> `

**Problem 6**

A projectile is fired from the point (0,0) with an intial speed of 80 meters per second and angle of elevation of

from the top of a hill that drops 1 meter for every 10 meters of horizontal distance.

Write parametric equations describing the projectile's position
*t*
seconds after the projectile was fired.

Plot the projectile's path for several values of
to estimate the value of
that causes the projectile to land as far away as possible.

**Hint**

**Hint #2**

Problem Setup - set equations and input variables
** **

`> `
**restart:**

`> `
**hilleq := -1/10*t:**

`> `
**bally := .5*g*t^2+voy*t+soy;**

`> `
**ballx := vox*t:**

`> `
**g := -9.8:
voy := 80*sin(theta):
soy := 0:
vox := 80*cos(theta):**

`> `

Start plotting

`> `
**with(plots):**

`> `
**hill := plot(hilleq,t=00..800,y=-100..200,color=green):**

`> `
**theta := Pi/4:
ang45 := plot([ballx,bally,t=0..15],color=red):**

`> `
**theta := Pi/5:
ang36 := plot([ballx,bally,t=0..15],color=blue):**

`> `
**theta := Pi/6:
ang30 := plot([ballx,bally,t=0..15],color=yellow):**

`> `
**theta := Pi/3:
ang60 := plot([ballx,bally,t=0..15],color=gold):**

`> `
**grade := evalf(solve(-1/10=tan(grade),grade));**

`> `
**angle := Pi/4+grade/2;**

`> `
**theta := angle:
longest := plot([ballx,bally,t=0..15],color=black):**

`> `
**display({ang45,ang36,ang30,ang60,longest,hill});**

To find the trajectory with the maximum distance, I found the grade of the hill and subtracted 1/2 it's value from 45*. That path is represented by the black line and the 45* path is represented by the red line. My black line calculation does go farther, but not by much.

`> `

`> `

`> `

`> `