Half of maximum possible projectile range

  • #1
kerbyjonsonjr
34
0

Homework Statement



A projectile's horizontal range on level ground is R=v02sin2[tex]\vartheta[/tex]/g. At what launch angle or angles will the projectile land at half of its maximum possible range

Homework Equations





The Attempt at a Solution



All I know is that a projectile goes the furthest when the angel is 45o. I know I have to say 1/2R = the whole thing but I am not sure what to do to solve this. There are two answers. Thanks for any help! I appreciate it.
 

Answers and Replies

  • #2
gneill
Mentor
20,945
2,886
Look at your formula for the range. What's special about 45° ?
 
  • #3
kerbyjonsonjr
34
0
Look at your formula for the range. What's special about 45° ?

It makes sin2x equal to 1. I GOT IT! I want the value where sin2x=.5 and thats at 15 and 75. Thank you!!!!!
 

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