Question

A particle is projected vertically upwards from top of the tower with velocity 10m/s .it reaches ground in 5s.gind distance transversed by the particle

Answer 1

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

intial 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 = s1 = ?

v²-u² = 2as1

s1 = v²-u²/2a

s1 = (0)²-(10)²/2(9.8)

s1 = 0-100/-19.6

s1 = -100/-19.6

s1 = 5.102m
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When we project the ball from tower then it reaches maximum hieght 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
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so total distance traveled by an object
S3 = s1 + s2

S3 = 245 + 5.102

S3 = 250.102 m

so total distance covered by an object is 250.102 meter
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If you want to find the height of the tower then ,
h = S2 - s1

h = 245 - 5.012

h = 239.898 meter
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Answer 2

Answer:

Explanation: see the attachment below.