Question

A gun is fitted on platform moving with velocity .v1= 10i^ m/s. a bullet is fired from the gun with muzzle velocity v2= (40 i^+20j^)m/s with respect to the platform. then the horizontal range of projectile is?

Answer 1

answer : option (3) 200m

explanation : velocity of platform,$v_p$= 10 i m/s

velocity of bullet with respect to platform, $v_{bp}$= (40 i + 20 j) m/s

so, velocity of bullet with respect to ground, $v_b=v_{bp}+v_p$

= 40i + 20j +10i

= 50i + 20j

velocity of bullet in horizontal direction, $u_x$ = 50m/s

velocity in vertical direction, $u_y$ = 20m/s

now, horizontal range, $R=\frac{2u_x.u_y}{g}$= {2(50)(20)}/10

= 200 m

hence, horizontal range = 200m

Answer 2

velocity of platform,$v_p$= 10 i m/s

velocity of bullet with respect to platform, $v_{bp}$= (40 i + 20 j) m/s

so, velocity of bullet with respect to ground, $v_b=v_{bp}+v_p$

= 40i + 20j +10i

= 50i + 20j

velocity of bullet in horizontal direction, $u_x$ = 50m/s

velocity in vertical direction, $u_y$ = 20m/s

now, horizontal range, $R=\frac{2u_x.u_y}{g}$= {2(50)(20)}/10

= 200 m

• hence, horizontal range = 200m