A balloon is at a height of 81 m and is ascending upwards with a velocity of 12 m-s- A body of 2kg weight is dropped from it- If g - 10 m-s2- the body will reach the surface of the earth in

H = ut + 1/2 gt^2 ⇒ 81 = 12t + 1/2× 10 × t^2 ⇒ t = 5.4 sec ...

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Standard XI

Physics

General Gravity Problem

A balloon is ...

Question

A balloon is at a height of 81 m and is ascending upwards with a velocity of $12 \frac{m}{s}$ . A body of 2kg weight is dropped from it. If $g = 10 \frac{m}{s^{2}}$ , the body will reach the surface of the earth in

1.5 s

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4.025 s

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5.4 s

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6.75 s

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Solution

The correct option is C 5.4 s
$h = u t + \frac{1}{2} g t^{2} \Rightarrow 81 = - 12 t + \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 5.4 s e c$

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A balloon is at a height of 81 m and is ascending upwards with a velocity of $12 \frac{m}{s}$ . A body of 2kg weight is dropped from it. If $g = 10 \frac{m}{s^{2}}$ , the body will reach the surface of the earth in

A. 4.025 s

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A balloon is at a height of 81 m and is ascending upwards with a velocity of 12ms. A body of 2kg weight is dropped from it. If g=10ms2, the body will reach the surface of the earth in

The correct option is C 5.4 sh=ut+12gt2⇒81=−12t+12×10×t2⇒t=5.4sec

A balloon is at a height of 81 m and is ascending upwards with a velocity of $12 \frac{m}{s}$ . A body of 2kg weight is dropped from it. If $g = 10 \frac{m}{s^{2}}$ , the body will reach the surface of the earth in

The correct option is C 5.4 s
$h = u t + \frac{1}{2} g t^{2} \Rightarrow 81 = - 12 t + \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 5.4 s e c$