Building Height Calculation Using Motion Equations

Question:

A particle is projected up with an initial speed of u=10 ms^-1 from the top of a building at time t=0.At time t=5 sec, the particle strikes the ground.Find the height of the building.

Solution:Building height calculation

Solution:

Net displacement of the particle is the difference of final position(C) and the initial position(A).

Applying the second equation of motion

$${s=ut+\frac{1}{2}at^2}$$

Here s=-H, a=-g=-10ms^-2 , u=+10 m/s

and t= 5 sec

$${-H=10(5)+\frac{1}{2}(-10)5^2}$$

$${-H=50-125}$$

$${-H=-75}$$

$${H=75 m}$$

Height of the building (H) =75 m