Query103Mth

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course Mth 277

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Question: Find the time of flight Tf (to the nearest tenth of a second) and the range Rf (to the nearest unit) of a projectile fired (in a vaccum) from ground level at `alpha = 65.54 degrees and v0 = 19.07 m/s. Assume that g = 9.8 m/s^2.

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Your solution:

r(t)=(v_0*cos(theta)ti+((v0*sin(theta)t-.5gt^2)j

r(t)=19.07m/s*cos(65.54)ti+[0+19.07m/s*sin(65.54)t-.5(9.8 m/s)t^2]

r(t)=(7.9 m/s)ti+[(17.4 m/s)t-(4.9 m/s)t^2]j

jcomp

[(17.4 m/s)t - (4.9 m/s)t^2]

t= 0, 3.6 s

i component=(7.9 m/s)t

The range of the projectile is 28.4 meters.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: An object is moving along the curve r = 1/(1 - sin(theta)), theta = t - pi/2. Find its velocity and acceleration in terms of the unit polar vectors u_r and u_theta.

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Your solution:

I extremely dislike this problem because of the mad chaos with coefficients.

r=1/(1-sin(theta))

dr/dt=dr/dTheta*dTheta/dt

dr/dt=cos(theta)/(1-sin(theta))^2*1

dr/dt=cos(t-pi/2)/(1-sin(t-pi/2))^2

This is where I got lost.

@&

It's not that chaotic once you get used to it, but it does look bad when you first see it. In any case, this is challenging.

Per the note I posted before this one, you also need r '' and theta ''. Since theta is linear, theta '' is zero.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: If a shotputter throws a shot from a height of 5.5t and an angle of 53 degrees with initial speed 28 ft/s. What is the horizontal distance of the throw?

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Your solution:

r(t)=(v_0cos(theta)ti+[h+v_0sin(theta)t-.5gt^2]j

5.5*t+28ft/s*sin(53)t-16ft/s^2*t^2=0

Quad Form

t=0,5

28ft/s*cos(53)*(5s)

85.2 ft

@&

That 5.5 t should have been 5.5 ft, but using 5.5 t your solution is correct. You know what you're doing here.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I felt confident with this question until my curious self went online and looked at the world record for shot put. The record is a little less than 76ft by american Randy Barnes. Unless this shotputter has superstrength then I believe I might have made an error in calculating something. I looked at my work for errors and could not find the mistake, so maybe you can see it.

@&

Starting from 5.5 ft the shot would take less than a second to come to rest, and less than a second to fall back down. With `dt less than 2 seconds, you get a more reasonable result.

We're not talking about a world-class shotputter here. Looks like the distance will be in the 40's. Not bad, but not particularly competitive.

*@

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Self-critique rating:OK

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Question: A child running along level ground at the top of a 40ft high vertical cliff at a speed of 15ft/s, throws a rock over the cliff into the sea below. If the rock is released 10 ft from the edge and at an angle of 45degrees, how long does it take the rock to hit the water and how far away from the base of the cliff does it hit?

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Your solution:

r(t)=(v_0cos(theta)ti+[h+v_0sin(theta)t-.5gt^2]j

v(t)=-v_0sin(theta)i+[v_0cos(theta)-gt]j

a(t)=-v_0cos(theta)i+[-v_0sin(theta) - g]j

r(t)=(v_0cos(theta)ti+[h+v_0sin(theta)t-.5gt^2]j

= [15 ft/s * cos(45)t]i+[40+15ft/s*sin(45)t-16ft/s^2*t^2]j

[40 + 15 ft/s*sin(45)t - 16ft/s^2*t^2] = 0

Quadr Form

One of the t's were neg so obviously not correct the other was t=1.947s

Range= 15 ft/s * cos(45)*1.947 s

Range= 20.65ft

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I solved this using physics after and got a similar answer, so unless I made two careless mistakes I believe my answer to be in the ballpark of the correct answer.

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Self-critique rating:OK

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Question: **A gun is fired with muzzle speed 700ft/s at an angle of 20degrees. It overshoots the target by 60 ft. If the target is moving away from the gun at a constant speed of 15ft/s and the gunner takes 30 seconds to reload, at what angle should the second shot be fired with the same muzzle speed?

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Your solution:

v0=700ft/s

theta=20degrees

[700ft/s*sin(20)*t+16t^2]j

t=15seconds

15ft/s*30s=450ft the target moves backward. Therefore, we need the bullet to travel 450ft-60ft=390ft farther.

Even though this isnt exact I added on .56s for the time for the bullet to travel the increased distance.

Found 7581ft by finding original distance and adding 390ft.

700ft/s*cos(theta)*15.56s=7581ft.

theta = 46 degrees

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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&#This looks good. See my notes. Let me know if you have any questions. &#