Question

a shell is fired from a Cannon at an angle of 30 degree with horizontal velocity of 1000 metre per second assuming G equal to 10 metre per second find time taken by the shell to reach the highest point and how high the shell will arise​

Answer 1

Answer:

ux=ucos60o=100cos60o

or ux=50m/s

At top must point the center of mass will have only horizontal component of velocity.

( & this will be ux=50 m/s)

If shell was of 3 m

then the fragment m will move upward with 200 m/s

so to make Net velocity of center of mass zero in vertical direction mx200+2mxvy=0

or vy=−100 m/s i.e downward.

only the bigger part is possessing the horizontal component of velocity so m+2mmx0+2mvx=ux

⇒Vx=23ux=75 m/s

speed=ux2+vy2=752+(−100)2=125 m/s