Question

An object is projected at such an angle(horizontal displacement) is three times the maximum height reached. The initial velocity of the object is 270m/s .Calculate the angle that the object is ppprojected at .

Answer 1

Answer:

we know , 

  horizontal range = u²sin2Ф/g

  maximum height = u²sin²Ф/2g

A/C to question,

    horizontal range = 3× maximum height 

    u²sin2Ф/g  =  3 × u² sin²Ф/2g 

   { sin2Ф = 2sinФ.cosФ }

    2sinФ.cosФ  = 3/2. sin²Ф

    tanФ = 4/3 

Ф = tan⁻¹(4/3) 

hope this helps

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