Question

a wooden block is dropped from the top of a building 100m high above ground simultaneously a bullet is fired from the foot of the building upwards witha velocity of 100m/s the height above the ground at which they meet is?

Answer 1

can anyone solve this

Answer 2

the ans=95.1m

Answer 3

Remember the kinematic equation? The bullet has position 0, initial velocity 100 m/s and acceleration -10m/s^2. The block has 0 initial velocity, initial position -100 and acceleration 10m/s^2. So:
100t - (1/2)(10)t^2 = 100 - (1/2)(10)t^2.
So: t = 1
Plug in on the bullet equation:
S = 100(1) - 5(1) = 95

Answer 4

let the height travelled by the wooden block be s1
s1=1/2*10*t^2
let the height travelled by the bullet be s2
s2=100t-1/2*10*t^2
both the cases time same
s1+s2=100
100=5t^2+100t-5t^2
100=100t
t=1
s1=5*1
=5m.
s2=100-5
=95m

Answer 5

There is a minor rounding due to the fact that I took g as 10 m/s^2 instead of 9.81 or something like that.