After some Google and asking Mr. GPT I found out that the return address is usually stored at the allocated stack size + 8 bytes on x86. So I disassembled assignment1c and found the following assembly code:

   0x0000000000401168 <+0>:	push   %rbp
   0x0000000000401169 <+1>:	mov    %rsp,%rbp
   0x000000000040116c <+4>:	sub    $0x90,%rsp
   0x0000000000401173 <+11>:	lea    -0x90(%rbp),%rax
   0x000000000040117a <+18>:	mov    %rax,%rdi
   0x000000000040117d <+21>:	call   0x401040 <gets@plt>
   0x0000000000401182 <+26>:	mov    0x8(%rbp),%rax
   0x0000000000401186 <+30>:	mov    %rax,-0x8(%rbp)
   0x000000000040118a <+34>:	mov    -0x8(%rbp),%rax
   0x000000000040118e <+38>:	lea    0xe7b(%rip),%rdx        # 0x402010
   0x0000000000401195 <+45>:	mov    %rax,%rsi
   0x0000000000401198 <+48>:	mov    %rdx,%rdi
   0x000000000040119b <+51>:	mov    $0x0,%eax
   0x00000000004011a0 <+56>:	call   0x401030 <printf@plt>
   0x00000000004011a5 <+61>:	nop
   0x00000000004011a6 <+62>:	leave

What we can see here is that with lea we allocate a size of 0x90 = 144 bytes. So with 144 + 8 being 152, I have to write 152 'A' characters and then put in my return address. So that's how the stack-c.sh file works.