from a cliff of 49m high ,a man drops a stone 1s later he throes second stone . they both hit the ground at the same time . find out the speed with which he throws the second stone.

1st Stone:

4.9t^2 = 49
t^2 = 10
Tf = 3.16 s. = Fall time.

2nd stone:
h = Vo*t + 0.5g*t^2 = 49
t = 3.16-1 = 2.16 s. = Fall time.

Vo*2.16 + 4.9*2.16^2 = 49
2.16Vo + 22.86 = 49
2.16Vo = 49-22.86 = 26.14
Vo = 12.1 m/s. = Initial velocity.

To find out the speed with which the man throws the second stone, we can start by calculating the time it takes for both stones to hit the ground.

Given:
Initial height (h) = 49m
Time delay between dropping the first and throwing the second stone = 1s

We can use the formula for the time it takes for an object to fall from rest to the ground:

t = √(2h / g)

where:
t is the time taken
h is the height
g is the acceleration due to gravity (approximately 9.8 m/s²)

First, let's calculate the time it takes for the first stone to hit the ground:

t₁ = √(2 * 49 / 9.8)
t₁ = √(98 / 9.8)
t₁ = √10
t₁ ≈ 3.16s

Now, let's calculate the time it takes for the second stone to hit the ground:

t₂ = t₁ + 1
t₂ ≈ 3.16 + 1
t₂ ≈ 4.16s

Since both stones hit the ground at the same time, the time taken for the second stone to hit the ground is also approximately 4.16 seconds.

Now, to find the speed with which he throws the second stone, we can use the formula for the speed of an object:

v = g * t

where:
v is the speed
g is the acceleration due to gravity
t is the time taken

Using the calculated time, we can determine the speed:

v = 9.8 * 4.16
v ≈ 40.77 m/s

Therefore, the speed with which the man throws the second stone is approximately 40.77 m/s.

To find the speed with which the man throws the second stone, we need to use the concept of free fall.

First, let's calculate the time it took for the first stone to hit the ground. We can use the formula for free fall height: h = (1/2) * g * t^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

Plugging in the values, we have:

49 = (1/2) * 9.8 * t^2

Simplifying the equation, we get:

t^2 = (49 * 2) / 9.8

t^2 = 10

Taking the square root of both sides, we find:

t ≈ √10 ≈ 3.16 seconds

Since the second stone was thrown 1 second after the first stone, it will take 3.16 - 1 = 2.16 seconds for the second stone to hit the ground.

Now, we can use the formula v = g * t to find the speed at which the second stone was thrown, where v is the velocity (or speed), g is the acceleration due to gravity, and t is the time taken.

Using the known values, we have:

v = 9.8 * 2.16

v ≈ 21.17 m/s

Therefore, the speed at which the man throws the second stone is approximately 21.17 m/s.