Grade 10Mechanics

a body is projected with a velocity of 40m/s After 2 sec it crosses a vertical pole of height 20.4m Find the angle of projection and horizontal range of projectile

Anupam

8 Years agoGrade 10

1 Answer

Arun

8 Years ago

U = 40 m/s. The height becomes 20.4 m after 2 sec.

h = u sinФ t - 1/2 g t²
20.4 = 40 sinФ * 2 - 1/2 * 9.8 * 2²
=> sinФ = 1/10 => Ф = sin⁻¹ 0.10
=> CosФ = √99 /10

=> sinФ ≈ 1/10

Hence sin2Ф = 2 sinФ cosФ = 2√99/100

= √99/50 = 9.949/50 = 0.19898 = 0.2

Horizontal range = u² sin2Ф / g ≈ 40² * 0.2/9.8
R = 32.49 m

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a body is projected with a velocity of 40m/s After 2 sec it crosses a vertical pole of height 20.4m Find the angle of projection and horizontal range of projectile

Anupam

1 Answer

Arun

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