A ball is thrown vertically upward with 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

For the 1st ball, the time of the upward motion
v=v₀-gt
v=0
0=v₀-gt
t= v₀/g =18/9.8=1.83 s.
The height of the 1st ball
h= v₀t-gt²/2=18•1.83 - 9.8•(1.83)²/2= =16.53 m
The time of the 1st ball downward motion before the 2nd ball begins to move is
Δt=2 s -1.83 s = 0.17 s
The 1st ball during 0.17 s covered the distance
Δh=g(Δt)²/2 =9.8•0.17²/2 = 0.14 m.
Its downward velocity is
v₀₁ = gΔt=9.8•0.17 =1.67 m/s
Now, two balls begin to move:
the 1st ball moves downward with initial velocity v₀₁ =1.67 m/s,
the 2nd ball moves upward with initial velocity v₀₂=13.5 m/s.
The distance separated them is
h₀ = h- Δh = 16.53 – 0.14 = 16.39 m.
Before the meeting, the 1st ball covered h₁=v₀₁t+gt²/2,
and the 2nd ball covered the distance h₂=v₀₂t-gt²/2.
h₀ = h₁+h₂=v₀₁t+gt²/2 + v₀₂t-gt²/2=
=(v₀₁+ v₀₂)t
t= h₀/(v₀₁+ v₀₂)=16.39/(13.5+1.67) =
=1.08 s.
The position above the ground is
h₂=v₀₂t-gt²/2 =13.5•1.08 – 9.8•1.08²/2 =
= 14.58 -5.72 = 8.86 m

Answer 2

A 1500 kg automobile is traveling up to 20 degree incline at a speed of 6 m/s. If the driver wishes to stop his car in a distance of 5m, determine the frictional force at pavement which must be supplied by rear wheels.

Answer 3

you can just solve to see when the heights are equal:

18t-4.9t^2 = 13.5(t-2)-4.9(t-2)^2
t = 1.08

proceed from there as above