hi vivek
lets just keep speed of the golf ball=v
and angle with hor.=A
hor. component of vel=v*cosA
vertical comp.=v*sinA
when A=45
cosA=sinA
so the velocities in hor. n vertical directions r eqal.
do the 2nd part urself...
cos37=4/5
sin37=3/5
the normal reaction here will be less than 50*9.8...bcoz the force F has a component in the vertical dir.
N=50*9.8 - 400*sin(20)
so ur eqn shud actually b
400N*Cos(20)-[(9.81 m/s^2)(50kg)-(400)(sin(20))](.60) = (50kg)(a)
i think u shud get the ans frm this.
bye
well its quite simple. ... take the vel of package in the backward hor. dir to b v=8.1+13=21.1m/s
H=16m
16=1/2*9.8*t*t evaluate t from this eqn.
d1=diatance traveled by the package in the hor. dir=v*t
d2=dist. traveled by the helicopter in the hor. dir.=8.1*t
hor. dist. bet the two...
angular velocity...(check it out n tell me what u guys think)
hello ppl...
consider this problem...
u could refer 2 resnick n halliday 5th edition, the 3rd author is krane. chapter 10, problem no. 8.
theres a disc that's rotating about a horizontal axis with angular velocity 'w'. a...