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# Numerical based on gravitation

Asked by raghavendran7 10th March 2010, 3:23 PM

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 Expert 10th March 2010, 7:02 PM
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