Correct answer: (c) 33.7 m

Explanation:

Assuming balls as point mass objects.

Let first ball is thrown at t = 0

And second ball will be thrown at t = 3s.

After 3s first ball will be at

S = u x 3 - \(\frac{1}{2}gt^2\)

S = 30 x 3 - \(\frac{1}{2}10 \times 9\)

S = 90 - 45 = 45 m

Velocity of ball (1) after 3s will be

v = 30 - 10 x 3 = 0

So it will be highest point.

Now let they collide after time t from the time when second ball is throwing and 1st ball is at height point.

S₁ = +\(\frac{1}{2}gt^2\)

S₂ = 30 x t - \(\frac{1}{2}10 \times t^2\)

30t = 45

t = 1.5 s

S₁ = \(\frac{1}{2}\)10 x 1.5 x 1.5

S₁ = \(\frac{15 x 1.5}{2}\)

S₁ = 11.25

S₂ = 33.75m

Nearly S₂ = 33.7m