Particle is projected up at 10m/s from the top of a 120m high tower Taking g=10m/s? Determine: (a) Time taken by it in reaching ground (b) Speed with which it hits the ground
Answers
Answer:
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=(0)²−(10)²/2(9.8)
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(5)+9.8(5)²
s
2
=0+9.8(25)
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