Question

A balloon rising vertically up with a uniform velocity 15 ms-1 releases a ball at a height of 50m. The time taken by the ball to hit the ground is (g = 10 ms-2 ).​

Answer 1

Answer:

Taking vertical downward motin of ball released from the balloon rising vertically upward, upto ground, we have ltbRgt ` u=-0 15 ms^(-1) , a = g = 10 ms^(-2)` , S = 100 m`,

` t= ?`

` S= ut + 1/2 at^2`

` 100 =- 15 xx t + 1/2 xx 10 xx t^2 =- 15 + 5 t^2`

or ` t^2 - 3 t- 20 = 0`

` or r^2 - 3 t- 20 =0`

or 1 t= ( 3 +- sqrt ( 9+ 80)/2 = (3 =- 9.4)/2 = 6. 2 s or ` -3.2 s`

As time cannot be negative, so ` t= 6.2 s`

Taking further upward motion of balloon from height ` 100o m` the heith attained by balloon in ` 6.2 s= 15 xx 6.2 = 93 .0` m

Total height of balloon ` = 100 + 93.0`

` = 193 m` .

Explanation:

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