Question

Two balls are thrown simultaneously ‘A’ vertically upward with a speed of 20 m/sec from the ground and ‘B’ vertically downwards from a height of 40 m with the same speed and along the same line of motion. At what point from the ground do the two balls collide? (Take g = 10 m/sec²)

1. 10 m
2. 15 m
3. 20 m
4. 25 m

Answer 1

Correct Answer - Option 2 : 15 m

CONCEPT:

• Equation of Kinematics: These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

V = U + at

\(s =ut+\frac{1}{2}{at^{2}}\)

V2 =U2+ 2as

Where u = Initial velocity of the particle at time t = 0 sec, v = Final velocity at time t sec, a = Acceleration of the particle, and s = Distance travelled in time t sec

CALCULATION:​

Given - Initial velocity (u) = 20 m/s

• Suppose the two balls meet at a height of x from the ground after time t sec.
• For the upward motion of ball A:-

\(\Rightarrow x\; = \;20t - \frac{1}{2} \times 10 \times {t^2}\; = \;20t - 5{t^2}\)      ------ (1)

• For the downward motion of ball B:-

\(\Rightarrow 40 - x\; = \;20t + \frac{1}{2} \times 10 \times {t^2}\; = \;20t + 5{t^2}\)     -------- (2)

On adding equation 1 and 2, we get

⇒ x + (40 - x) = 20t – 5t² + 20t + 5t²

⇒ t = 1 sec

On substituting the value of t = 1 sec in equation 1, we get

 So,  x = 20 × 1 – 5 × 1² = 15 m