Posted by **Seetharaman** on Monday, July 22, 2013 at 11:54am.

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?

- BTech, Mechanical -
**Elena**, Monday, July 22, 2013 at 1:44pm
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

- BTech, Mechanical -
**Seetharaman**, Monday, July 22, 2013 at 2:54pm
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.

- BTech, Mechanical -
**Steve**, Monday, July 22, 2013 at 3:43pm
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

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