a projectile is thrown with speed u = 20 m/s at an angle 30 degree with horizontal from the top of building 40 m high. the horizontal range of projectile is
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u = 20 m/s. Ф = 30°. To find x when y = - 40 m.
Equation of projectile: y = x tan Ф - g x² Sec² Ф / 2u²
y = -40 m when the projectile hits the ground.
y = - 40 = x /√3 - (10 * 4/3 * 1/2 * 1/20² ) x²
x² - 20√3 x - 2400 = 0
x = 10√3 + √2700 = 40 √3 meters
Range = 40 √3 m
Equation of projectile: y = x tan Ф - g x² Sec² Ф / 2u²
y = -40 m when the projectile hits the ground.
y = - 40 = x /√3 - (10 * 4/3 * 1/2 * 1/20² ) x²
x² - 20√3 x - 2400 = 0
x = 10√3 + √2700 = 40 √3 meters
Range = 40 √3 m
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Answer:
40√3
Explanation:
see the attachment
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