Question

"A shot is fired from a point at a distance of 200 m from the
foot of a tower 100 m high so that it just passes over it
Find the magnitude and direction of the velocity of shot

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Answer 1

For a projectile in 2-dimensions, we have the path equation as:

y = x tanФ - gx² sec²Ф /2u² 

        u = initial speed,         Ф = angle of projection

        x = horizontal displacement,    y = vertical displacement

We assume that the tower height 100 m is the maximum height achieved by the bullet.  Then the distance 200m will be equal to half of the range of the bullet.

 

We know :   H = u² Sin²Ф/2g 

               100 = u²  Sin²Ф /(2*10)

=>     u² sin²Ф = 2,000  --- (1)

                  R = u² Sin2Ф/g

           2*200 = u² Sin2Ф/10

=>   u² Sin2Ф = 4,000     --- (2)

(2) ÷ (1) =>     TanФ = 1   =>  Ф = 45°

              =>     u² = 4000

                       u = 20√10 m/s

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