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A balloon moves up vertically such that if a stone is thrown from it with a horizontal velocity v relative to the balloon, it always hits the ground at a fixed point 2(v^2)/g horizontally away from it(the balloon). Find the height of the balloon as a function of time.

A balloon moves up vertically such that if a stone is thrown from it with a horizontal velocity v relative to the balloon, it always hits the ground at a fixed point 2(v^2)/g horizontally away from it(the balloon). Find the height of the balloon as a function of time.

Grade:10

1 Answers

Arun
25750 Points
6 years ago
The height of the balloon is constant with respect to time. As the range is given to be constant every time, the height has to be constant, since the range will depend upon the time taken by stone to reach the ground , which depends upon height of balloon.

Let the velocity of the balloon when it is at a height h  be v. Hence when the stone is thrown from the balloon with a velocity u of  in the horizontal direction, relative to the balloon, the actual velocity of the stone, with respect to the ground will be u  in the horizontal direction and v in the vertical. Now, they say that irrespective of the height of the balloon above the ground the range of the stone is 2u^2/g. Hence if the balloon takes a time t0  to reach the ground then we can easily write .

ut0=2u^2/g

t0=2u^2/ug=2u/g

Also using the second equation of motion in the vertical direction, we obtain . Substitute the value of  to obtain an equation of the form

h=-vt0+1/2gt^2

substituing the value of t0

h=-v(2u/g)+1/2gt^2

v+gh/2u=u

dh/dt+gh/2u=u

This is standard differential equation with soltuin h(exp)(gt/2u)=ut+C

where C is constant or integration

h=ut+Cexp)(-gt/2u)

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