Classic problem

Well, I wanted to use calculus and I did!:) I supposed the block is somewhere in its journey, and has compressed by a distance ' $x$ '. Remember, this $x$ is not the maximum compression, but some arbitrary compression at the moment it's velocity is maximum. So, by energy conservation,

$\quad \quad \quad mg(h+x)\quad =\quad \frac { 1 }{ 2 } k{ x }^{ 2 }+\frac { 1 }{ 2 } m{ v }^{ 2 }\\ \Rightarrow \quad 60(5+x)\quad =\quad { x }^{ 2 }+3{ v }^{ 2 }\\ \Rightarrow \quad { v }^{ 2 }\quad =\quad 100+20x-\frac { 1 }{ 3 } { x }^{ 2 }\\ \Rightarrow \quad v\quad =\quad { (100+20x-\frac { 1 }{ 3 } { x }^{ 2 }) }^{ \frac { 1 }{ 2 } }$

Now, by differentiating and equating to $0$ , we get $x=30$ and using this values $v=20$

Cheers!:)

If you are thinking that answer should be 10 then you are wrong.

As soon as block touches the spring, spring will surely force it upwards but that force will be not enough to retard the block until spring force becomes equal to blocks weight $i.e. mg$ . So spring will compress upto $x=\frac{mg}{k}$ untill the block starts to retard.

Now apply energy conservation.

You should get $v=\sqrt{2gh +\frac{mg^2}{k}}$

Plug in the values

you may also use the fact that in shm the velocity is maximum at mean position.

The essential fact is after the block hits the spring with K.E. = P.E. = $mgh$ it still accelerates because initially the retarding force $kx$ on it is NOT equal to it , infact it is less than it till mg = kx at $x$ $=$ $30.0$ $m$ . But the motion between x = 0 and x = 30 m is quite complicated as acceleration varies as $a(t)$ $=$ $ω^2$ $x(t)$ . So to bypass this we use energy considerations.

$mgh$ $+$ $mgx$ $=$ $0.5kx^2$ $+$ $0.5mv^2$ putting values we get $v$ $=$ $20$ $ms^{-1}$ .

(p.s. this problem reminds me of an ipho problem involving bungee jumping.i think it was IPHO 2002/2001.)