Question

A stone is dropped from a balloon going up with a uniform velocity of 10 m/s. If the height of the balloon was 56.25 m when the stone was dropped, then the time after which the stone will strike the ground after it is dropped is [

Answer 1

h = ut + 1/2*g*t^2
Now,
initial velocity of the balloon as well as the stone will be - 10 m/s.
So, therefore,
56.25 = -10t + 1/2*10*t^2
5t^2 - 10t - 56.25 = 0
t = 10 +-√ 4*5*56.25 / 10 = 10+-√1125/10 = 10+-15√5 /10
Now,
10-15√5 / 10 will be a negative value and hence we'll neglect because negative time obviously makes no sense.

Thus, the actual time taken will be
10+15√5 / 10 seconds.

Answer 2

Hope u like my process

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Formula to be used

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$=> S = \frac{v^{2 } - u^2} {2g} \\ \\ => S = ut + \frac{1}{2} gt^2$

where

=>S = distance travelled.

=>u = initial velocity

=> g = gravity = 10m/s² (given)

=> t = time taken

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Since the balloon was going up with a velocity (u = 10 m/s)

During the droppage it must have moved a little more towards upward.

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So.. The more it goes upward is

$=> x = \frac{v^{2 } - u^2}{2g} =\frac{0^{2 } - 10^2}{2(-10)}\\ \\ \: \: \: x = \frac{-100} {-20} =\bf \underline{5 \: \: m} \\ \\ time \: \: taken(t_{1}) = \frac{v-u} {g} =\frac{-10} {-10} =\bf \underline{ 1 \: \: \: sec}$

So..

Total height attained by stone

= 56.25 +5 = 61.25 m

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Now..

After being at top height..

=> Initial velocity = u = 0 m/s

=> S = height = 61.25 m

=> g = gravity =10 m/s²

So.. Let Time taken will be given by t sec.

$=> S = ut +\frac{1 }{2} gt^2\\ \\ => 61.25 = 0\times t + \frac{1} {2}\times 10t^2\\ \\ => 61.25 = 5 t^2 \\ \\ => t = \sqrt{\frac{61.25}{5}}= \bf \underline{3.5 \: \: \:sec}$

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So time taken to strike the ground will be = $t +t_{1 }$ = 3.5 + 1 = 4.5 sec.

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