Question

You stand at the top of a tall building with a stopwatch. You drop a rock off the side of the building, and it takes the rock 1.9 seconds to hit the ground. (Assume no air resistance.) How tall is the building? What was the impact speed of the rock? If you were standing at the bottom of this building, at what speed would you have to throw the rock for it to reach the top of the building?

Answer 1

Given,

Time taken by the rock to fall to the ground, t=1.9 s

As you are dropping the rock, the initial velocity, u=0 m/s

Acceleration of the rock is equal to the acceleration due to gravity, g=9.8 m/s²

To find the height of the building:

From the equation of motion, we have

`h=ut+(1)/(2)gt^2`

Where h is the height of the building.

On substituting the known values,

`h=0+(1)/(2)*9.8*1.9^2=17.69\text{ m}`

To find impact speed:

From the equation of motion,

`v=u+at`

Where v is the impact velocity.

On substituting the known values,

`v=0+9.8*1.9=18.62\text{ m/s}`

To find the velocity with which the rock needs to be thrown:

Here, we need to find initial velocity u.

When the rock is thrown, it will stop when it reaches the maximum height. Let us assume the maximum height is equal to the height of the building.

Therefore the velocity of the rock when it reaches the top of the building, i.e., the final velocity, v=0 m/s

As the rock is moving against gravity, the acceleration of the rock will be, -g=-9.8 m/s²

From the equation of motion,

`v^2-u^2=-2gh`

On substituting the known values,

`0-u^2=-2*9.8*17.69=-346.72`

Therefore the initial velocity is,

`u=\sqrt[]{346.72}=18.62\text{ m/s}`

Therefore, the height of the building is 17.69 m

and the impact velocity is 18.62 m/s

The initial velocity with which the rock needs to be thrown is 18.62 m/s