Question

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. The angle of projection O is​

Answer 1

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 α

  VCosα * 2T  = 2x

=> VCosα * T  =  x

=> VCosα = x/T

=> x = VCosα.T

for vertical

S = √3x

t = (2T/2) = T

a = -g

Velocity at top = 0

V = u + at

0 = VSinα - gT

=> VSinα = gT

S = ut + (1/2)at²

√3x =  VSinα * T + (1/2)(-g)T²

=> √3x =  VSinα * T - (1/2)(gT).T

Substitute gT = VSinα , x = VCosα.T

=> √3VCosα.T = VSinα * T - (1/2)VSinαT

cancel VT

=>  √3Cosα = Sinα  -  (1/2)Sinα

=> √3Cosα =  (1/2)Sinα

=> 2√3 = Sinα/Cosα

=> tanα = 2√3

=> α = tan⁻¹(2√3)

=> α = 73.9°

The angle of projection is 73.9°

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Answer 2

For subjective exams, please follow the above procedure given by Amitnrw sir. I will tell you a easier method to solve it as an MCQ (especially in NEET and JEE- Mains).

---------------_----------------_----------------------

• Let the angle of elevation of the highest point of the projectile with respect to the point of projection be $\alpha$.

• Let angle of elevation of the highest point of the projectile with respect to the point at which the projectile again touches the ground be $\beta$

• Let angle of projection be $\theta$.

[Kindly remember the following expression , use it only in MCQs]

$\therefore \: \tan( \theta) = \tan( \alpha ) + \tan( \beta )$

• Now , both $\alpha \:and\: \beta$ are 60° because the projectile starts at one base vertex and ends at the other base vertex.

Putting the values:

$\implies \: \tan( \theta) = \tan( {60}^{ \circ} ) + \tan( {60}^{ \circ} )$

$\implies \: \tan( \theta) = \sqrt{3} + \sqrt{3}$

$\implies \: \tan( \theta) = 2\sqrt{3}$

$\implies \: \theta \approx \: {73.89}^{ \circ}$

So, angle of projection is approx 73.89°.