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Physics 17 Online
OpenStudy (anonymous):

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)

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

OpenStudy (anonymous):

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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

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

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

OpenStudy (shamim):

h=ut+(1/2)at^2 490=0*t+(1/2)*9.8t^2 490=4.9t^2 t^2=100 t=10

OpenStudy (anonymous):

.-.

OpenStudy (shamim):

the above calculation is for vertical component

OpenStudy (shamim):

now for horizontal component s=vt

OpenStudy (anonymous):

yes

OpenStudy (shamim):

v=360km/h convert it into m/s

OpenStudy (shamim):

t=10s

OpenStudy (shamim):

v=360*1000m/3600s

OpenStudy (anonymous):

yup got itr

OpenStudy (anonymous):

o_o you should memorise the formula

OpenStudy (anonymous):

yes i learn't projectile.......... just yesterday

OpenStudy (shamim):

anyway i slways enjoy open study

OpenStudy (anonymous):

i see

OpenStudy (dan815):

omgosh xD i forgot it was in km

OpenStudy (dan815):

sry

OpenStudy (dan815):

360 km/h= x m/s , x=360*1000/60^2 = 100 m/s therefore 100m/s * 10s = 1000m

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