Figure shows a projectile thrown with speed u= $20$ m/s at an angle $30$ with horizontal from the top of a building $40$ m high then the horizontal range of projectile is.

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Solution

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

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Figure shows a projectile thrown with speed u=20 m/s at an angle 30 with horizontal from the top of a building 40 m high then the horizontal range of projectile is.

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

Figure shows a projectile thrown with speed u= $20$ m/s at an angle $30$ with horizontal from the top of a building $40$ m high then the horizontal range of projectile is.

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