Question

ONE SECOND after ball A is projected upwards with a velocity of 16 m/s, a second ball, B, is thrown vertically downwards at a velocity of 9 m/s from a balcony 30 m above the ground.

1.Calculate how high above the ground ball A will be at the instant the two balls pass each other?

Answer 1

The distance covered by the 1st ball for 1 second is
hₒ=vₒ₁t - gtₒ²/2 =16•1 -9.8•1²/2 = 16 -4.9 = 11.1 m (above the ground)
Its speed at this height is
v= vₒ₁ - gtₒ=16-9.8•1=6.2 m/s.

h=H-hₒ=30-11.1=18.9 m
Now to balls separated by distance ‘h’ are moving towards each other

h₁=vt- gt²/2,
h₂=vₒ₂t + gt²/2,
h= h₁+h₂=vt- gt²/2+ vₒ₂t + gt²/2 = vt+ vₒ₂t = (v+ vₒ₂)t.

t= h/(v+ vₒ₂)=18.9/(6.2+9) = 18/9/15.2=1.24 s.

h₁=vt- gt²/2 =6.2•1.24 -9.8•1.24²/2= =7.688 -7.534=0.154 m
The height above the ground is eaqual to hₒ+h₁=
= 11.1+0.154 = 11.254 m

Answer 2

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