An aeroplane is flying horizontally at a height of 490m with a velocity of 360km/h. A bag containing ration is to be dropped to the soldiers on the ground. How far from the ground should the bag be released so that it falls directly over them.(g=9.8m/s^2)
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OpenStudy (dan815):
ok lets do this
OpenStudy (anonymous):
yes
OpenStudy (dan815):
okay so calclate the time it takes to fall first
OpenStudy (dan815):
Vy(0)=0,
Y(0)=450,
so
Y=1/2*at^2+450
0=1/2g*t^2+450
solve for t
OpenStudy (dan815):
and then do 360*t, and thats how far u must drop it away
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OpenStudy (anonymous):
|dw:1402480943298:dw|
OpenStudy (dan815):
g=-'ve not positibe
OpenStudy (anonymous):
Can we use the range formula somehow
\[\huge R=u.\sqrt{\frac{ 2H }{ g }}\]
OpenStudy (dan815):
well that is kinda wha we doing
OpenStudy (anonymous):
Ok so, let's follow the formula
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OpenStudy (dan815):
0=1/2g*t^2+450
t=sqrt(2*450/9.8)
OpenStudy (anonymous):
after multiplying by 360 i ain't gettingthe answer , i 'm getting
3449.9
answer is 1000
OpenStudy (dan815):
490*
OpenStudy (dan815):
the anwer is 3600 horizontal distance away from the target
OpenStudy (anonymous):
it is given 1000
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OpenStudy (shamim):
1000 m is right answer. i calculated it in my mind
OpenStudy (anonymous):
how
OpenStudy (dan815):
no mitake
OpenStudy (dan815):
d^y/dt^2=-9.8
dy/dt=-9.8t+k1
y=-4.9t^2+k1t+k2
y'(0)=0=k1
y(0)=490=k2
y=-4.9t^2+490
@Y being height t where height =0
y=0=-4.9t^2+490
y=sqrt(490/4.9)
=sqrt100
360*sqrt100 = 3600