A ballon is ascending at the rate `v = 12 km//h` and is being carried horizontally by the wind at `v_(w) = 20 km//h`. If a ballast bag is dropped from the balloon at the instant h = 50 m, determine the time needed for it to strike the ground. Assume that the bag was released from the ballon with the same velocity as the balloon. Also, find the speed with which the bag strikes the ground?

Answer

B, C

Explanation

`u_x = 20 km//h = 20 xx 5/18 = 5.6 m//s`
`u_y = 12 km//h`
`= 12 xx 5/18 = 3.3 m//s`
Using `S = ut + 1/2 at^2` in vertical direction, we have
`-50 = (3.3) t + 1/2 (-10) t^2`
Solving this equaiton we get,
`t = 3.55 s`
At the time of striking with ground,
`v_x = u_x = 5.6 m//s`
`v_y = u_y + a_yt`
`= (3.3) + (-10) (3.55)`
`=32.2 m//s`
`:. Speed = (sqrt(32.2)^2 + (5.6)^2)`
` ~~ 32.7 m//s`.