Question

A shot is fired from a point at a distance of 200m from the foot of a towre 100m hight so that it just passes over it. find the magnitude and diretion of velocity of the shot

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

Answer 2

Answer:

Here is the answer

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