Question

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)?

a) When will the bolt and the apple be at the same height? (while they are both in the air)?
b) At what height above the ground will the two objects meet?

Answer 1

use 'relative' concept to solve easily. look at the explanation.
Read it repeatedly until U get clear.

Explanation:

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)