Question

From the foot of a tower 90m high stone is thrown up so as to reach the top of a tower two seconds later another stone is dropped from height of Tower with two stones
meet at

Answer 1

Height of the building (d) = 90m

Let’s find the velocity of stone thrown up:

We know, v^2 – u^2= 2ad

Velocity of stone after reaching top of the building = 0

Acceleration is opposite to acceleration due to gravity = -9.8m/s^2

0- u^2= 2*(-9.8)*90

- u^2=-1764

Therefore, velocity of stone moving up= 42m/s

Distance at time t = h(t) =ut-1/2at^2

h(t) = 42t-4.9t^2 -----(1)

h1(T) = height of building – 1/2aT^2

T=t-2 (since second stone is thrown after two seconds)

H1= 90 – 4.9(t-2)^2

When the two stones meet, distance are same. So, Set h= h1

42t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.6

42t-19.6t=90-19.6

22.4t = 70.4 => t = 3.14s

Substitute the value of “t” in equation (1)

                           h =83.6m