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Projectile Motion PLEASE HELP MIDTERM TOMMOROW

  • #1
Projectile Motion PLEASE HELP MIDTERM TOMMOROW!!!

1. A ball is launched with a velocity of 1000m/s at some angle theta. The ball must travel 2000m in the x dir and land 800m in the y dir on a plateau. What theta should the ball be shot at to make this happen.



2. y=y0 +v0t -.5gt^2



3. I solved for t using x=v0cos(theta)t and plugged that into the y equation and got one solution. The answer key says theres two solutions. One at 20 something, which is the one i got and also one at 89.4 degrees. Can someone show me how to get the other/both solutions


THANKS!!!!!! I appreciate it
 

Answers and Replies

  • #2
10
0
You should use the formula
y=x tan theta - [tex]\stackrel{gx^2}{2u^2cos^2 theta}[/tex]

THis formula is combined by
verticial component using s=ut+1/2 at^2
horizontal component v=s/t
 
  • #3
i used that equation and still only got 22.4 degrees at the sole solution.... 89.4 doesnt seem to fit. Is there an error in the answer key?
 
  • #4
10
0
i also get the 22.4 degree
but it is acceptable that there is another angle is the answer.....
however I forget how to solve it ~sorry
 
  • #5
ok thanks anyway!
 

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