Question

A bomb is projected at an angle 60 degree with the horizontal speed 20 metre per second.At the hightest point it explodes into three parts of equal masses .If just after explosion one part comes to rest and other part retraces its path, then the distance of third part from poiny of projection is

Answer 1

Answer:

Explanation though it explodes at the highest point particle travelling anywhere among that path will continue moving in that direction in which was initially moving in respective of the other particles stopping or unable to continue journey

Thus by simply using the range formulae we can find out the distance travelled

Range=(velocity)^2x(sin2x)/g

Where x is the angle of inclination

g is acc due to gravity =10m/s

Velocity is the speed with which the particle is thrown

Range=12^2xsin(2x60)/10

Range/distance= 24.9=30m

Sin(120)= sin(180-60)=sin(60)=✓3/2

Answer 2

Answer:

√3 / 2 is answer

May it help u !!