Question

A stone is dropped from the roof of a building, 1.10s after that a second stone is thrown straight down with an initial speed of 20.0 m/s & the two stones land at the same time

How high is the building h=???m

Answer 1

Suppose Height of build is h,

Now if 1st stone takes time t to reach the ground, then second stone takes time "t - 1.10"

Now for first stone:

U = initial velocity = 0 m/sec

time taken to reach the ground = t

a = acceleration due to gravity = g = 9.81 m/sec^2

Using 2nd kinematic equation:

h = U*t + (1/2)*a*t^2

h = 0*t + (1/2)*9.81*t^2

Now for second stone:

U = initial velocity = 20 m/sec

time taken to reach the ground = t - 1.10

a = acceleration due to gravity = g = 9.81 m/sec^2

Using 2nd kinematic equation:

h = U*t + (1/2)*a*t^2

h = 20*(t - 1.10) + (1/2)*9.81*(t - 1.10)^2

So solving above two equation, Since height of building is same,

0*t + (1/2)*9.81*t^2 = 20*(t - 1.10) + (1/2)*9.81*(t - 1.10)^2

Since (a - b)^2 = a^2 - 2*a*b + b^2

(1/2)*9.81*t^2 = 20*t - 20*1.10 + (1/2)*9.81*t^2 - (1/2)*9.81*2*1.10*t + (1/2)*9.81*1.10^2

0 = 20*t - 20*1.10 - (1/2)*9.81*2*1.10*t + (1/2)*9.81*1.10^2

(20 - 9.81*1.10)*t = 20*1.10 - (1/2)*9.81*1.10^2

9.209*t = 16.06495

t = 16.06495/9.209

t = 1.744 sec

So Using this value of time

h = (1/2)*9.81*1.744^2

h = 14.91 m = height of building

to check the answer, solve 2nd equation too

h = 20*(t - 1.10) + (1/2)*9.81*(t - 1.10)^2

h = 20*(1.744 - 1.10) + (1/2)*9.81*(1.744 - 1.10)^2 = 14.91 m

So, Height of building = 14.91 m

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