A projectile moving vertically upwards with a velocity of 200 ~ms ⁻¹ breaks into two equal parts at a height of 490 ~m . One part starts moving vertically upwards with a velocity of 400 ~ms ⁻¹ . How much time it will take, after the break up with the other part to hit the ground?

Step-by-step solution

Momentum before explosion = Momentum after explosion gathered m 200 j = m 2 400 j + m 2 v = m 2 (400 j +v) 400 j -400 j =v v=0 gathered i.e., the velocity of the other part of the mass, v=0 Let time taken to reach the earth by this part be t Applying formula, h=u t+ 1 2 g t^2 aligned & 490=0+ 1 2 9.8 t^2 & t^2= 980 9.8 =100 & t= 100 =10 sec aligned