ABC is a equilateral triangle in a vertical plane. From point A a projectile is projected such that it just Grazes B and lands at C. The angle of projection theta is
Answers
Given : ABC is an equilateral triangle in a vertical plane. From point A a projectile is projected such that it just grazes Band lands at C.
To Find : The angle of projection
Solution:
Assume that length of Equilateral triangle = 2x ( AB = BC = AC = 2x)
Then Height of Triangle = √((2x)² - x²) = √3x
Let say after 2T time it reaches C then time T to reach mid way at point B.
Projected with speed V at angle θ
VCosθ * 2T = 2x
=> VCosθ * T = x
=> VCosθ = x/T
=> x = VCosθ.T
for vertical
S = √3x
t = (2T/2) = T
a = -g
Velocity at top = 0
V = u + at
0 = VSinθ - gT
=> VSinθ = gT
S = ut + (1/2)at²
√3x = VSinθ * T + (1/2)(-g)T²
=> √3x = VSinθ * T - (1/2)(gT).T
Substitute gT = VSinθ , x = VCosθ.T
=> √3VCosθ.T = VSinθ * T - (1/2)VSinθT
cancel VT
=> √3Cosθ = Sinθ - (1/2)Sinθ
=> √3Cosθ = (1/2)Sinθ
=> 2√3 = Sinθ/Cosθ
=> tanθ = 2√3
=> θ = tan⁻¹(2√3)
=> θ = 73.9°
The angle of projection is tan⁻¹(2√3) or 73.9°
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