Stone moved in horizontal circle of radius 1.5m by string height of 2m above ground.the spring break and particle flies off horizontally striking ground 10m away.find centripetal acceleration ?

To find the centripetal acceleration of the stone, we can use the formula:

a = v^2 / r

Where:
a = centripetal acceleration
v = velocity of the stone
r = radius of the circle

First, we need to find the velocity of the stone. We can do this by using the equation for projectile motion:

s = ut + (1/2)at^2

Where:
s = horizontal distance traveled by the stone (10m)
u = initial velocity of the stone (0 m/s)
a = acceleration due to gravity (9.8 m/s^2)
t = time taken for the stone to hit the ground

Since the stone is moving in a horizontal circle, the time taken for it to hit the ground is the same as the time taken for one complete revolution. We can find the time taken for one complete revolution using the formula:

T = 2πr / v

Where:
T = time taken for one complete revolution
r = radius of the circle
v = velocity of the stone

Substitute the values:
T = 2π(1.5) / v

Now, substitute the value of T into the equation for projectile motion:

10 = 0 + (1/2)(9.8)(T)^2

Solve for T and then substitute it back into the equation for T to find the velocity of the stone.

Once you have the velocity of the stone, you can then calculate the centripetal acceleration using the formula a = v^2 / r.

Given that ,
r = 1.5 m
h = 2 m
l = 10 m

By the conservation of energy ,
loss in potential energy = gain in kinetic energy
mgh =1/2×mv^2
v^2 = 2gh
v = √(2×9.8×2)
= √39.2
v =6.26 ms^-1

So , centripetal acceleration ,
a_c = v^2/r
= (6.26)^2/1.5
=25.40 m/s^2