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

Answer 1

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 $=s=?$

$v²-u²=2as$

$s=v²-u²/2a$

$s=\left(0\right)²-\left(10\right)²/2\left(9.8\right)$

$s=0-100/-19.6$

$s=-100/-19.6$

$s=5.102m$

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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 $=s_2=?$

$s_2=ut+at²/2$

$s_2=0\left(5\right)+9.8\left(5\right)²$

$s_2=0+9.8\left(25\right)$

$s_2=245m$

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so total distance traveled by an object

$S_3=s_1+s_2$

$S_3=245+5.102$

$S_3=250.102m$

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 ,

$h=S_2-s_1$

$h=245-5.012$

$h=239.898$ meter