Question

A ball is thrown vertically upward with a velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. at what position above the ground will they meet?

Answer 1

The time of the first ball motion to the top point is
v=v(o1) –gt, v=0,
t=v(o1) /g=18/9.8 = 1.84 s.
The maximum height of the 1st ball is
H= v(o1) •t –gt²/2=16.53 m.
When the 2nd ball starts, the 1st ball is on its return trip during 2-1.84 =0.16 s.
It covers the distance hₒ=g(0.16)²/2 =0.125 m. and has the downward velocity
v(1)=g(0.16)=1.568 m/s.
At this instant the distance between two balls is 16.53-0.125=16.4 m, and two balls are moving towards each other: the first ball accelerates downward with initial velocity v(1)=1,568 m/s, the second ball decelerates upward at initial velocity v(2)=13.5 m/s. They cover the distances:
the1st ball h1= v1•t+gt²/2; the 2nd ball h2=v2•t-gt²/2.
h1+h2= 16.4 m.
=> v1•t+gt²/2+ v2•t-gt²/2 =16.4
t=16.4/(v1+v2)=16.4/(1.568+13.5)=1.088 s.
The point where the balls mett is located
h= v2•t-gt²/2 =13.5•1.088-g•(1.088)²/2 =8.89 m (above the ground)