1s 5 m. 5 m

Sol. u = m/s.

∴ u cos60° = m/s

and u sin60 0 =

= 7.5 m/s

Since the horizontal displacement of both the shots are equal , the second should be fired early because its horizontal component of velocity u cos60 0 is less than the other’s which is u or m/s.

Now let first shot takes t 1 time to reach the point P and the second t 2 . Then –

x = ( u cos 60º ) t 2 = u. t 1

or x = t 2 = t 1 ....(i)

or t 2 = 2t 1 ....(ii)

and h = g = – (7.5) t 2

Taking g = 10 m/s 2

h = 5t 2 2 – 7.5 t 2 = 5t 1 2 ....(iii)

Substituting t 2 = 2t 1 in equation (3), we get

5( 2t 1 ) 2 – 7.5 (2 t 1 ) = 5t 1 2

or 5t 1 2 = 5t 1

t 1 = 0 and 1s

Hence t 1 = 1s and

t 2 = 2t 1 = 2s

x = t 1 = m ( From equation i )

and h = 5 t 1 2 = 5 (1) 2 = 5 m ( From equation iii )

∴ y = 10 – h = ( 10 – 5 ) = 5 m

Hence

(i) Time interval between the firings = t 2 – t 1 = ( 2 – 1) s

Δ = 1s

(ii) Coordinates of point = (x, y ) = m. 5 m