Half of maximum possible projectile range

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To determine the launch angles for a projectile to land at half of its maximum range, the formula R = v0^2 sin(2θ) / g is used. The maximum range occurs at a 45° angle, where sin(2θ) equals 1. To find the angles for half the maximum range, sin(2θ) must equal 0.5, which occurs at angles of 15° and 75°. This means that launching a projectile at either of these angles will result in it landing at half of its maximum possible range. Understanding the relationship between the launch angle and the sine function is crucial for solving this problem.
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Homework Statement



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

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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.
 
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Look at your formula for the range. What's special about 45° ?
 
gneill said:
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 that's at 15 and 75. Thank you!
 

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