CBSE Class 9 Answered
Numerical based on gravitation
Asked by raghavendran7 | 10 Mar, 2010, 03:23: PM
Expert Answer
Dear Student
Time of flight
1)
t =2 X V/g
==> v=tg/2=6X10/2=25.85 m/sec
2)V=(2gh)
h=v^2/2g=33.4m
3) At the end of third second projectile researches it s amx height
Distance travelled with initial velocity o and acceleration g
is s=0.5 gt^2=0.5 mt
So projectile is 33.4-5=28.4 m from the ground.
Regards
Team
Topperlearning.com
Answered by | 10 Mar, 2010, 07:02: PM
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