Acompletely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. just before the collision, one ball of mass 3.0 kg is moving upward at 20 m/s, and the other ball, of mass 2.0 kg, is moving downward at 14 m/s. how high do the combined two balls of putty rise above the collision point? (neglect air drag.)

h = 2.087 m

Explanation:

Given

m₁ = 3 kg

v₁ = 20 m/s

m₂ = 2 kg

v₂ = - 14 m/s

In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.

In the given problem, lets assume that the balls of putty are initially moving along the  y  axis, upward direction being the positive  y  direction. And the collision occurs at the origin of the coordinate system.

We can apply the equation

vs = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)  

⇒   vs = (3 kg*20 m/s + 2 kg*(- 14 m/s)) / (3 kg + 2 kg)  

⇒   vs = 6.4 m/s (↑)

To calculate the maximum height  h  attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity:

vf² = vi² - 2*g*h

where

vf = 0 m/s  

g = 9.81 m/s²

vi = vs = 6.4 m/s

finally we get h:

h = vi² / (2*g)

⇒   h = (6.4 m/s)² / (2*9.81 m/s²) = 2.087 m

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min