Question

A boy on a 20 m high cliff drops a stone. One second later, he throws down another stone. Both the stones hit the ground simultaneously. Find the initial velocity of the second stone.

Answer 1

height of cliff , h = 20m

Let time taken by the dropped stone is t

initial velocity of dropped stone , u = 0

so, s = ut + 1/2 at²

-h = 0 - 1/2gt²

t = √{2h/g} = √{2 × 20/10} = 2 sec

a/c to question, one second later,boy throws down another stone . so, second stone must be reach the ground in 1 sec.

so, s = ut + 1/2at²

here, h = 20m/s , t = 1sec and g = 10m/s²

now, 20 = u + 1/2 × 10 × 1²

20 = u + 5

u = 15 m/s

hence, initial velocity of second stone is 15 m/s downward direction.

Answer 2

Answer:

15 m/s

Step-by-step explanation:

Consider the first stone falls freely under gravity.

Hence, Initial Velocity (u) = 0.

Height of the cliff (H) = 20m.

Acceleration = g

From the second equation of motion,
s = ut + ½gt²

⇒20 = ½(10)t²

⇒20 = 5t²

⇒t² = 4

⇒t = 2

After one second the first stone is dropped, The boy drops his second stone.

But the question says, Both of the stones reached the same time.

So, Time taken by the second time to reach ground = 1 second.

As the stone took less time, We must think the boy dropped with some Velocity.

Let that Velocity be u.

Now, Applying in laws of motion

20 = u(1) + 1/2(10)(1²)

20 = u + 5

20 - 5 = u

u = 15m/s

Therefore, Initial Velocity of the second stone is 15 m/s.