Question

A body of mass M is thrown vertically upwards with a velocity U it reaches to some height and returns to ground after sometimes the total change in momentum taking place is 6 MG kilometre per second total time of flight of the body is given by g is the acceleration due to gravity

Answer 1

So here comes the concept of time of acent = time of decent

As the ball is expected to come back in 4 sec. i.e from starting point to the heighest and then to the starting point it takes 4 sec. So time to reach highest point will be 2 sec.

Now using kinematics

v=u+a*t

Initial velocity u is unknown and as the velocity at the highest point is always 0 so

u=-(-10)2

u=20mps.

Now

s=u*t+(1/2)*a*t^2

S=20*3+(1/2)*(-10)(3)^2

S= 60–45

s=15m. Above ground

So here comes the concept of time of acent = time of decent

As the ball is expected to come back in 4 sec. i.e from starting point to the heighest and then to the starting point it takes 4 sec. So time to reach highest point will be 2 sec.

Now using kinematics

v=u+a*t

Initial velocity u is unknown and as the velocity at the highest point is always 0 so

u=-(-10)2

u=20mps.

Now

s=u*t+(1/2)*a*t^2

S=20*3+(1/2)*(-10)(3)^2

S= 60–45

s=15m. Above ground