Question

13. A ball dropped from the top of tower falls first hal
height of tower in 10 s. The total time spend by ba
in air is [Take g = 10 m/s2]
1 14.14 s
(2) 15.25 s
(3) 12.36 s
(4) 17.36 s​

Answer 1

Correct Question -

A ball dropped from the top of the tower falls first half height of the tower in 10s. The total time spent by a ball in the air is [take g = 10 m/s²].

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Answer -

Given -

h = h/2

u = 0 m/s

t = 10 s

a = g = 10 m/s²

where

$\longrightarrow$h is height of tower.

$\longrightarrow$t is time taken to cover distance half of tower.

$\longrightarrow$a is acceleration due to gravity.

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To find -

Total time - T

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Formula used -

$\implies$S = ut + 1/2 at²

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Solution -

Using the 2nd equation of motion -

$\bf s = ut + 1/2 gt^2$

$\implies$$\bf \frac{h}{2} = \frac{1}{2} \times 10 \times 10 ^{2}$

h = 1000 m

Height of tower = 1000 m

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$\longrightarrow$h = 1000

$\longrightarrow$u = 0

$\longrightarrow$a = g = 10 m/s²

Substituting the value in 2nd equation of motion -

$\bf s = ut + 1/2 gT^2$

$\implies$$\bf 1000 = \frac{1}{2} \times 10 \times {T}^{2}$

$\implies$$\bf {T}^{2} = 200$

$\implies$$\bf T = 10 \sqrt{2}$

$\implies$$\bf T = 14.14 sec$

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Thanks

Answer 2

Question :

• A ball dropped from the top of tower falls first half .The height of tower in 10 s. The total time spend by ball in air is [Take g = 10 m/s²]

Solution :

As per the given question ,

The ball is dropped from the top of the tower it means that the initial velocity (u)of the ball is zero

>> u = 0 m /s

According to the question the ball covers half the height of the tower in 10 seconds

>> t = 10 s

Let the height of the tower be H

As the ball uniformly accelerated in the presence of gravity we can use the second kinematics equation in order to solve this question

we know that ,

$\bigstar \: \boxed{\mathtt{H = ut + \dfrac{1}{2} g{t}^{2} }}$

here ,

• H= height of the tower
• u = initial velocity
• t = time taken
• g = acceleration due to gravity

Substituting the given values in the above equation we get ,

$\rightarrow\mathtt{ \dfrac{H}{2} = \dfrac{1}{2} g {t}^{2} }$

$\rightarrow\mathtt{ H= g {t}^{2} }$

$\rightarrow\mathtt{ H= 10 \times 100}$

$\rightarrow\mathtt{ H= 1000 \: m}$

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we need to find the total time spent by the ball in air

By using the second kinematic equation ,

$\bigstar \: \boxed{\mathtt{H = ut + \dfrac{1}{2} g {t}^{2} }}</p><p>$

Substituting the given values in the above equation we get ,

$\rightarrow\mathtt{H = \dfrac{1}{2} g {t}^{2} }$

$\rightarrow\mathtt{t = \sqrt{ \dfrac{2H}{g} } }$

$\rightarrow\mathtt{t = \sqrt{ \dfrac{2 \times 100 \cancel0}{\cancel{10}} } }$

$\rightarrow\mathtt{t = \sqrt{200} }$

$\rightarrow\mathtt{t = 10 \sqrt{2} }$

$\rightarrow\mathtt{t = 10 \times 1.414}$

$\rightarrow\mathtt \red{t = 14.14 \: s}$

The total time spent by the ball in air is 14 .14 m