#$&* course Mth 277 query_10_3*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique rating:OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK "