Question

5. A balloon starts rising from ground from rest with an upward acceleration 2 m/s2. Just after 1 s, a stone is dropped from it. The time taken by stone to strike the ground is nearly ​


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Answer 1

Answer:

EXPLANATION.

A balloon starts rising from ground from rest

with an acceleration = 2 m/s².

After, 1 seconds a stone is dropped from it.

To find time taken by the stone to strike

the ground is nearly.

$\sf : \implies \: a \: = 2ms {}^{ - 2} \\ \\ \sf : \implies \: balloon \: starts \: from \: rest \: = u = 0 \\ \\ \sf : \implies \: time \: taken \: = 1 \: seconds \\ \\ \sf : \implies \: \green{{ \underline{from \: newton \: 1 {}^{st}equation \: of \: kinematics }}} \\ \\ \sf : \implies \: v \: = u + at \\ \\ \sf : \implies \: v \: = 0 \: + \: 2 \times 1 = 2ms {}^{ - 1}$

$\sf : \implies \: distance \: travelled \: by \: balloon \: in \: 1 \: seconds \\ \\ \sf : \implies \: \green{{ \underline{from \: newton \: 3 {}^{rd} equation \: of \: kinematics}}} \\ \\ \sf : \implies \: {v}^{2} = {u}^{2} + 2as \\ \\ \sf : \implies \: s \: = \frac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \sf : \implies \: s \: = \frac{4 - 0}{4} = 1 \\ \\ \sf : \implies \: s \: = 1 \: m$

$\sf : \implies \: the \: acceleration \: of \: stone \: after \: dropped \: is \: \\ \\ \sf : \implies \: u \: = - 2ms {}^{ - 1} \\ \\ \sf : \implies \: a \: = 9.8ms {}^{2} \\ \\ \sf : \implies \: s \: = 1 \: m \: \\ \\ \sf : \implies \: t \: = 1 \: seconds$

$\sf : \implies \: \green{{ \underline{from \: newton \: 2 {}^{nd}equation \: of \: kinematics }}} \\ \\ \sf : \implies \: s \: = ut \: + \: \frac{1}{2} a {t}^{2} \\ \\ \sf : \implies \: 1 = - 2t \: + \: \frac{1}{2} \times 9.8 \times {t}^{2} \\ \\ \sf : \implies \: 4.9 {t}^{2} - 2t \: - 1 = 0$

$\sf : \implies \: factorise \: in \: the \: form \: of \: quadratic \: equation \\ \\ \sf : \implies \: x \: = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \sf : \implies \: t \: = \frac{2 \pm \: \sqrt{( - 2) {}^{2} - 4 \times 4.9 \times ( - 1) } }{2 \times 9.8} \\ \\ \\ \sf : \implies \: t \: = \frac{2 \pm \: \sqrt{4 + 19.6} }{2 \times 9.8} \\ \\ \sf : \implies \: t \: = \frac{2 \pm \: \sqrt{23.6} }{9.8} \\ \\ \sf : \implies \: t \: = 0.7 \: \: seconds$

$\sf : \implies \: \green{ { \underline{time \: taken \: by \: stone \: to \: strike \: the \: ground \: nearly \: = 0.7 \: seconds}}}$

Answer 2

When the stone is dropped from the balloon, its initial velocity is the same as the velocity of the balloon at that instant.

 The upward acceleration of the balloon is a=2m/s2

 The balloon starts from rest, so u= 0 m/s

 The balloon rises for 1 s before the stone is dropped, hence we have

 v =u + at

v=0+=2 m/s in the vertically upw ard direction.

 This is the initial velocity of the stone after being dropped.

 The distance moved up by the balloon in 1 second is

v2=u2+2as

v2=2as

∴S=2av2=44=1m

 Hence, the stone falls by 1 m before hitting the ground

 Now, the acceleration on the stone after being dropped is g=9.8m/s2

 Hence, for the stone, 

we have 

u=−2m/s;s=1m;a=9.8m/s2;t=1s

u is negative as it is in the upward direction

 Hence, using third equation of motion,

 we get time as

s=ut+21at2

1=−2t+21×9.8t2

4.9t2−2t−1=0

∴=

t=2×9.82−+(−2)2−4×4.9×(−1)=

2×9.82−+4+19.6=0.7seconds---answer