A balloon starts rising from the surface of the earth with vertical component of velocity `v_(0)`. The balloon gathers a horizontal velocity `v_(x) = ay`, where a is a constant and y is the height from the surface of the earth, due to a horizontal wind. Determine
(a) the equation of trajectory of the balloon.
(b) the tangential, normal and total acceleration of the balloon as function of y.

A balloon starts rising from the surface of the earth with a verticle component of velocity v_(0) . The baloon gathes a horizontal velocity v_(x)=lambda_(y) , where lambda is a constant and y is the height from the surface of earth, due to a horizontaln wind. determine (a) the equation of trajectory and (b) the tangential, normal and total acceleration of the balloon as a function of y .

A balloon starts rising from the surface of the Earth. The ascertion rate is constant and equal to v_0 . Due to the wind the balloon

A balloon starts rising from the earth's surface. The ascension rate is constant and equal to v_(0) . Due to the wind. The balloon gathers the horizontal velocity component v_(x)=ky , where k is a constnat and y is the height of ascent. Find how the following quantities depednd on the height of ascent. (a) the horizontal drift of the balloon x (y) (b) the total tangential and normal accelrations of the balloon.

A balloon starts rising from the earth's surface. The ascension rate is constant and equal to v_(0) . Due to the wind. The balloon gathers the horizontal velocity component v_(x)=ky , where k is a constnat and y is the height of ascent. Find how the following quantities depednd on the height of ascent. (a) the horizontal drift of the balloon x (y) (b) the total tangential and normal accelrations of the balloon.

A stone is dropped from a balloon rising with acceleration a. The acceleration of the stone relative to the balloon is :

Two ballons are simultaneously released from two buildings A and B. Balloon from A rises with constant velocity 10ms^(-1) While the other one rises with constant velocity of 20ms^(-1) Due to wind the balloons gather horizontal velocity V_(x)=0.5 y, where 'y' is the height from the point of release. The buildings are at a distance of 250 m & after some time 't' the balloons collide.

A ball is thrown vertically upward with a velocity u from balloon descending with a constant velocity v. The ball will pass by the balloon after time