A weight is dropped from the top of a building $135$ metre high. How fast is the weight moving just before it hits the ground?

Open in App

Solution

Formula,

$v^{2} = u^{2} + 2 a s$

$= 0 + 2 \times 10 \times 135$

$∴ v = 232.4$

Suggest Corrections

Similar questions

Q. A tv tower is 40m high.A mass of

5 k g

is dropped from its top . calculate the potential energy at the top and kinetic energy just before it hits the ground

(

take

g = 10 m / s^{2})

Q. The velocity of a toy dropped from a 20 m tower just before it hits the ground is: (Take g = 10)

Q. A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground? (g = 10 m/s²)

A ball is thrown vertically upwards with a velocity of 20 metre per second from the top of a multi storey building .The height of the point from where the ball is thrown is 25 m from the ground .

How high will the ball rise and how long will it be before the ball hits the ground ?

Take g = 10 m/s

I got X - X₀as 20 m . Is this the distance ball rised or is that X ?

Thank 😺 You

Q. A stone is dropped from the top of a building 147 m high. How long will it take to reach the ground? What will be its velocity when it strikes the ground?

Join BYJU'S Learning Program

A weight is dropped from the top of a building 135 metre high. How fast is the weight moving just before it hits the ground?

Formula,v2=u2+2as=0+2×10×135∴v=232.4

A weight is dropped from the top of a building $135$ metre high. How fast is the weight moving just before it hits the ground?

Formula,

$v^{2} = u^{2} + 2 a s$

$= 0 + 2 \times 10 \times 135$

$∴ v = 232.4$