An object is released from an aeroplane which is diving at angle of 30 degrees to the horizontal with a speed of 50 m/s. If the plane is at a height of 4000m from the ground when the object is released find
(a) the velocity of the object when it hits the ground
(b) the time taken by object to reach the ground
take g=9.81 m/s^2
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Verified answer
vy(t)=-sin(30)*50-9.81*t
and
y(t)=4000-sin(30)*50*t-.5*9.81*t^2
when y(t)=0, the object strikes the ground.
0=4000-25*t-.5*9.81*t^2
t=26.12 s at impact
vy(26.12)=-25-9.81*26.12
vy=-281 m/s
and vx=cos(30)*50
vx=43.3 m/s
the resultant v is
sqrt(vy^2+vx^2)
=284.6 m/s
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