Q. A balloon is moving up with a speed of 60 m/s.When the balloon is at a height 225 m above the ground, a stone is thrown horizontally with speed 40 m/s with respect to the balloon.

Find:

a) The horizontal displacement of the stone when it is at the top of its trajectory.

b) The speed of the stone when it hits the ground.

c)The horizontal displacement of the stone when it hits the ground.

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The relative velocity of the stone will be obtained from a resultant of two velocities, one horizontal, 40m/s and the other vertical, 60 m/s. Applying triangle law of vector addition,

Velocity of throwing, v = sqrt(60^2+40^2) = 72.11 m/s

and the angle of throwing (with the horizontal), theta = (90°-(arctan(40/60))*180/pi) = 56.3°

At the top of its trajectory, the vertical component of velocity is zero. Applying laws of motion for this component,

0=vsintheta-gt

`rArr` t = vsintheta/g =(72.11*sin56.3°)/9.81=6.1155 s

a) Horizontal range covered by the stone in this time =

vcostheta*t=(72.11*cos56.3°)*6.1155 =244.7 m

Total height attained = (225+72.11*sin56.3°*6.1155) m=591.9 m

b) Speed of the stone when it reaches the ground = sqrt(2*9.81*591.9) m/s = 107.76 m/s

c) Time taken to descend = sqrt(2*591.9/9.81)=10.9851 s

horizontal range covered by the time of descent =

10.9851*72.11*sin56.3° = 439.52 m

Total horizontal range covered during its entire flight =

439.52+244.7=684.2 m.

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