Question
Wed March 10, 2010 By: Raghavendran C.v

Numerical based on gravitation

Expert Reply
Wed March 10, 2010

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

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