Projectile motion range

  • Thread starter Cole07
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  • #1
Cole07
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Question: How far will a stone travel over level ground if it is thrown upward at an angle of 31.0 degrees with respect to the horizontal and with a speed of 11.0m/s?


What is the maximum range that could be achieved with the same initial speed?


I have tried to solve this with no sucess what i have done is Drawn a triangle with a angle being 31.0 degrees the opposite side from the angle is Vy the adjacent side is Vx the hypothesis is 11.0m/s then i drew a chart with X and Y X side of the chart I have Vx= 10.06216594 a=0 then on the y side i have
Vy=-4.444414099 a=-9.8 the Vf=0m/s I used the formula Vf=Vi+At and solved for time and got
.453116428 then i used d=Vi*t+1/2At^2 and solved for D and got 2.132070497 for the first question (But was told this was not correct)

For the second question i doubled the time and multiplied by Vx D=10.06216594 (0.906232856) and got 9.118665377 but it was wrong !

Is there anyone who might be able to help me Please???
 

Answers and Replies

  • #2
drpizza
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lmao, your calculator is in radians mode. Switch it to degrees mode :wink: After you have the time, don't forget it goes up and comes back down...

Your horizontal component will give you the distance, after you know the total time in the air.

45 degrees is a nice angle to maximize distance.
 
Last edited:
  • #3
Cole07
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Do you have to double time to solve for D with the D=ViT+1/2At^2 I know you must double it to get the maximum range (I am reworking the problem in degree mode I can't believe I did that!)
 
  • #4
Cole07
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Is the horzontial component 11.0cos(31.0) =9.428840308 this is the distance?
 

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