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

Explanation

From equation of motion
v. `= u + at ` ….(1)
`h = ut + 1/2 at^2`
` v. = av, u = v , a = + g`
using eqn (1)
3v = v + gt `rArr` 3v - v = gt
` t =(2v)/(g)`
substituting .t. value in eqn (2)
`h = v ( (2v)/(g) ) + 1/2 g ( (2v)/(g) )^2 = (2v^2)/(g) + 1/2 g ( (2v)/(g) )^2`
` h = (2v^2)/(g) + (2v^2)/(g)`
` = (4v^2)/(g) " " g = 10 ms^(-2)`
` = (4v^2)/(10) , h = (2v^2)/(5)`