A ball falls from height h. After 1 second, another ball falls freely from a point 20 m below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h :- (g = 10 m/s)

Question

A ball falls from height h. After 1 second, another ball falls freely from a point 20 m below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h :- (g = 10 m/s) (A) 11.2 m (B) 21.2 m (C) 31.2 m (D) 41.2 m

Tardigrade Admin · Accepted Answer

For first ball h =(1/2) gt 2 ...(i) For second ball ( h -20)=(1/2) g ( t -1)2 ...(ii) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7db8c260b467a6a0feb529594b6fd7b4-.png" /> subtract equation (ii) from (i) 20=-5+10 t ⇒ ∴ t =2.5 sec Hence, height h =(1/2) × 10 ×(2.5)2=31.2 m

Q. A ball falls from height $h$ . After $1$ second, another ball falls freely from a point $20 m$ below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h :- $(g = 10 m / s)$

11.2 m

21.2 m

31.2 m

41.2 m

For first ball $h = \frac{1}{2} g t^{2}$ ...(i)
For second ball $(h - 20) = \frac{1}{2} g (t - 1)^{2}$ ...(ii)

subtract equation (ii) from (i)
$20 = - 5 + 10 t \Rightarrow ∴ t = 2.5 sec$
Hence, height $h = \frac{1}{2} \times 10 \times (2.5)^{2} = 31.2 m$

Q. A ball falls from height h. After 1 second, another ball falls freely from a point 20m below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h :- (g=10m/s)