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Question

A particle is projected vertically up from the top of a tower with velocity 10m/s. It reaches the ground in 5s . Find-
a) Height of tower
b)Striking velocity of particle at ground
c)Distance traversed by particle
d)Average speed & average velocity of particle

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Solution

When we project the ball from the tower then it's

initial velocity =u=10m/s

final velocity=v=0m/s (when it reaches maximum height)

acceleration due to gravity=a=g=9.8m/s²(here - sign because object is moving opposite direction of acceleration)

distance =s=?

v²u²=2as

s=v²u²/2a

s=(0)²(10)²/2(9.8)

s=0100/19.6

s=100/19.6

s=5.102m
____________________________________________________________

When we project the ball from tower then it reaches maximum height then it will be in rest at that moment.Then it starts to fall down

intial velocity = u = 0m/s

acceleration due to gravity=a=g=9.8m/s²

time =t=5s

distance =s2=?

s2=ut+at²/2

s2=0(5)+9.8(5)²

s2=0+9.8(25)

s2=245m
_________________________________________

so total distance traveled by an object
S3=s1+s2

S3=245+5.102

S3=250.102m

so total distance covered by an object is 250.102 meter
___________________________________________
If you want to find the height of the tower then ,
h=S2s1

h=2455.012

h=239.898 meter


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