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"Angle Of Inclination" problem
A small ball is launched from the corner of an inclined plane. When the inclination of the plane α is 70deg the maximum range of the projectile is 50cm. Later, the inclination of the plane changes to new unknown angle, and the same projectile launcher is oriented at 30 deg to the horizontal line. The ball lands 1m away from the launcher on the inclined plane. Find the new angle of inclination.
R = v2sin(2x) / g
NB: For those of you who haven't heard of this, you attached a ball launcher to the corner of a piece of wide wood (or other solid material), and you incline said solid material at different angles. As you go down, the ball travels further horizontally on the plane. It had something to do with altering g at different angles of inclination
For the first situation, I know that the degree of the launcher was 45o, since we're referring to maximum range.
For the second situation, I had the idea of doing sin(2*45) / sin(2*30) = 1.1547. This might be totally wrong, but my "idea" at the time was that I could find the ratio with which the max. range was greater than the given range (if that makes any sense)
Since 1.1547 * 1 = 1.1547, I can do 1.1547 / 0.5 = 2.3094, giving me the ratio of the new angle of inclination to the old one.
Henceforth, 70 / 2.3094 = 30.3 degrees (approximately)
Honestly, did I approach this problem in the right way at all? I feel like that I did something horribly wrong when trying to solve it...
Homework Statement
A small ball is launched from the corner of an inclined plane. When the inclination of the plane α is 70deg the maximum range of the projectile is 50cm. Later, the inclination of the plane changes to new unknown angle, and the same projectile launcher is oriented at 30 deg to the horizontal line. The ball lands 1m away from the launcher on the inclined plane. Find the new angle of inclination.
Homework Equations
R = v2sin(2x) / g
The Attempt at a Solution
NB: For those of you who haven't heard of this, you attached a ball launcher to the corner of a piece of wide wood (or other solid material), and you incline said solid material at different angles. As you go down, the ball travels further horizontally on the plane. It had something to do with altering g at different angles of inclination
For the first situation, I know that the degree of the launcher was 45o, since we're referring to maximum range.
For the second situation, I had the idea of doing sin(2*45) / sin(2*30) = 1.1547. This might be totally wrong, but my "idea" at the time was that I could find the ratio with which the max. range was greater than the given range (if that makes any sense)
Since 1.1547 * 1 = 1.1547, I can do 1.1547 / 0.5 = 2.3094, giving me the ratio of the new angle of inclination to the old one.
Henceforth, 70 / 2.3094 = 30.3 degrees (approximately)
Honestly, did I approach this problem in the right way at all? I feel like that I did something horribly wrong when trying to solve it...
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