At a construction site, a bolt slips over the edge of a steel girder, which is 35 m above the ground. At the same instant, an apple is thrown vertically upwards from the ground with a speed of 15 m/s. How do you solve parts a) and b)?

since both are under gravity, their relative acceleration is zero.
now

At time t=0, #U_(ba)# {i.e., velocity of bolt(b) with respect to apple(a) } is the initial relative velocity of ball+apple system, and their initial relative separation, #s_o#, is 35m.
so using 2nd eqn of motion, #s=s_o + ut+1/2a t^2#, when they meet at time t=t, their relative distance becomes zero i.e.,#s=0#

Using this, #0=35-15t +1/2 (0)t^2#
{here U need to notice one thing that I've used ' - ' before 15t
this is coz I've assumed the origin to be at ground with Y axis up. so, #s_o# is kept ' + ', while #U_(ba)#' - 'as this vector is pointing downward i.e., in negative y direction}. Don't worry, It's very asy to understand, Just feel what's happening.
now
(A) t=35/15 or, 2.33 sec.

(B) Let me solve this part using bolt (U can choose either).
look, in same reference frame as mentioned earlier, final position of bolt from origin(i.e., ground),s, is given by
#s=35+0*(2.33)+1/2*(-9.8)*(2.33^2)#
solve this to get s.(here also I use -9.8 coz 'g' is pointed in negative Y axis.)

( Direction matters a lot in vectors)