Question

A body is projected at an angle of projection of 35° with horizontal. to get the same range with the same velocity of the projection , the body can also be projected at an angle of. A) 55° B) 65° C) 70°. D) 80°​

Answer 1

Answer:

Explanation:

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)