Question:

A stone is thrown upwards with a speed v from the top of a tower.It reaches the ground with a speed 3v.What is the height of the tower?

Solution:

Thrown up – Sign convention

Tower height calculation:

Applying third equation of motion

$${v^2=u^2+2as}$$

Here initial velocity u=+v

Final velocity v= -3v

Acceleration due to gravity a=-g

dispacement s=-h

Substituting these in the equation of motion

$${(-3v)^2=v^2+2(-g)(-h)}$$

$${9v^2=v^2+2gh}$$

$${h=\frac{8v^2}{2g}}$$

$${h=\frac{4v^2}{g}}$$

Height of the tower is   $${h=\frac{4v^2}{g} \ m}$$

Tower height calculation-image

Result:

The height of the tower $${h=\frac{4v^2}{g} \ metre}$$