A projectile of mass 3m explodes at highest pt of its path.It breaks into three equal parts.One part retraces its path, the second one comes to rest.The range of the projectile was 100m if no explosion would have taken place.The distance of the third part from the pt of projection when it finally lands on the ground is
Answers
When the projectile reaches the highest point, its vertical component of the velocity disappears altogether.
Only this is the horizontal component of velocity residue.
Therefore, before the explosion, Momentum = 3 M × U × Kos ......................... (1)
It has broken into three equal parts. A part of the mass to retract its path.
Therefore, after the explosion, its speed is the angle in the horizontal direction.
The second part of the mass comes to rest. Therefore its velocity is zero.
Let u move the remaining third part of the speed
According to the preservation of speed, 3 meters × u × cosα = m × u × cosα + m × 0 + m × u .............. (2)
eqn (2), we get u = 2 × u × kos ............................. (3)
Therefore, after the explosion in the third part, there is 2 × u × cosα velocity in the horizontal direction, as well as it falls due to gravity.
Time to reach the highest point for the original mass of 3 meters = u × sinα / g ................ (4)
The time taken for the third part of the mass is also available for reaching the ground from the highest point as given in the econ. (4)
Therefore, the horizontal distance is traveled by the third part of the mass = 2 × u × ko × × × sinα / g = u2 × sin2α / g .................. (5 )
The RHS (5) of EKN is similar to the horizontal range of the projectile, which is given as 100 meters.
Therefore, the third part of the mass will cover the distance of 100 meters after the explosion.
Half range before the explosion distance traveled by the original mass i.e. 50 meters.
Therefore, the required landing distance of the third part of the mass from the projection point is 100 + 50 = 150 meters.