Question

please solve question 35
with explanations
answer -> A​

Answer 1

Q. From the top of a tower a stone is thrown up and reaches the ground in time $t_1[/tex<strong>]a second stone is thrown down with the same speed and reaches the ground in time</strong>[tex] t_2$and a third stone release and reaches the ground in time $t_3$.

Answer:-

$T_3 = \dfrac{(T_1 + T_2)}{2}$

Option → A

Given :-

A stone is thrown upward and reaches the ground in time $t_1$.

A 2nd stone is thrown with same u and reaches the ground at $t_2$.

A 3rd stone is released from rest and reaches the ground at $t_3$.

To find :-

The relation between $t_1 , t_2 , t_3$

Solution:-

Let the initial velocity of stone be u and final velocity be v.

• We used here 3 cases.

Case :- 1

When stone is thrown upward.

u = u

v = 0 m/s

T = t'

Then ,

Time taken to reach maximum height will be:-

→$v = u -gt$

→$0 = u-gt'$

→$gt' = u$

→$t' = \dfrac{u}{g}$

Now,

ball reaches at max. height so, it's initial velocity will be 0.

u = 0 m/s

v = v

T = t"

Time taken to reach the ground:-

→$v = u+gt"$

→$v = 0 + gt"$

→$v = gt"$

→$t" = \dfrac{v}{g}$

Since, Time taken to reaches the stone in ground is $T_1$

→$T_1 = t' + t"$

→$T_1 = \dfrac{u}{g} + \dfrac{v}{g}$

→$T_1 = \dfrac{u + v }{g}$

Case :- 2

When stone is thrown downward it initial velocity is same that is in previous case.

u = u

v = v

$T = T_2$

Time taken to reach the ground is :-

→$v = u+ gT_2$

Case :- 3

When ball is released.

u = 0 m/s

v = v

→$T = T_3$

→$v = u+ gT_3$

→$v = gT_3$

According to question,

Both cases has same u.

→$T_1 = \dfrac{u + v}{g}$

• put the value of u.

→$T_1= \dfrac{v-gT_2+v}{g}$

→$T_1 = \dfrac{2v -gT_2 }{g}$

• Put the value of v.

→$T_1 = \dfrac{2(gT_3)-gT_2}{g}$

→ $T_1 = \dfrac{2gT_3-gT_2}{g}$

→$T_1 = \dfrac{g(2T_3-T_2)}{g}$

• cancel out g.

→$T_1 = 2T_3-T_2$

→$T_1 + T_2 = 2T_3$

→$T_3 = \dfrac{(T_1+T_2)}{2}$

hence,

The relation between them is :-

$T_3 = \dfrac{(T_1 + T_2)}{2}$

Answer 2

Answer:

Here is your answer

thrown up and reaches the ground in time t_1[/tex]a second stone is thrown down with the same speed and reaches the ground in time$t_2t </p><p>1</p><p> </p><p>$asecondstoneisthrowndownwiththesamespeedandreachesthegroundintime[tex]t

2

and a third stone release and reaches the ground in time t_3t

3

.

Answer:-

T_3 = \dfrac{(T_1 + T_2)}{2}T

3

=

2

(T

1

+T

2

)

Option → A

Given :-

A stone is thrown upward and reaches the ground in time t_1t

1

.

A 2nd stone is thrown with same u and reaches the ground at t_2t

2

.

A 3rd stone is released from rest and reaches the ground at t_3t

3

.

To find :-

The relation between t_1 , t_2 , t_3t

1

,t

2

,t

3

Solution:-

Let the initial velocity of stone be u and final velocity be v.

We used here 3 cases.

Case :- 1

When stone is thrown upward.

u = u

v = 0 m/s

T = t'

Then ,

Time taken to reach maximum height will be:-

→v = u -gtv=u−gt

→0 = u-gt'0=u−gt

′

→gt' = ugt

′

=u

→t' = \dfrac{u}{g}t

′

=

g

u

Now,

ball reaches at max. height so, it's initial velocity will be 0.

u = 0 m/s

v = v

T = t"

Time taken to reach the ground:-

→v = u+gt"v=u+gt"

→v = 0 + gt"v=0+gt"

→v = gt"v=gt"

→t" = \dfrac{v}{g}t"=

g

v

Since, Time taken to reaches the stone in ground is T_1T

1

→T_1 = t' + t"T

1

=t

′

+t"

→T_1 = \dfrac{u}{g} + \dfrac{v}{g}T

1

=

g

u

+

g

v

→T_1 = \dfrac{u + v }{g}T

1

=

g

u+v

Case :- 2

When stone is thrown downward it initial velocity is same that is in previous case.

u = u

v = v

T = T_2T=T

2

Time taken to reach the ground is :-

→v = u+ gT_2v=u+gT

2

Case :- 3

When ball is released.

u = 0 m/s

v = v

→T = T_3T=T

3

→v = u+ gT_3v=u+gT

3

→v = gT_3v=gT

3

According to question,

Both cases has same u.

→T_1 = \dfrac{u + v}{g}T

1

=

g

u+v

put the value of u.

→T_1= \dfrac{v-gT_2+v}{g}T

1

=

g

v−gT

2

+v

→T_1 = \dfrac{2v -gT_2 }{g}T

1

=

g

2v−gT

2

Put the value of v.

→T_1 = \dfrac{2(gT_3)-gT_2}{g}T

1

=

g

2(gT

3

)−gT

2

→ T_1 = \dfrac{2gT_3-gT_2}{g}T

1

=

g

2gT

3

−gT

2

→T_1 = \dfrac{g(2T_3-T_2)}{g}T

1

=

g

g(2T

3

−T

2

)

cancel out g.

→T_1 = 2T_3-T_2T

1

=2T

3

−T

2

→T_1 + T_2 = 2T_3T

1

+T

2

=2T

3

→T_3 = \dfrac{(T_1+T_2)}{2}T

3

=

2

(T

1

+T

2

)

hence,

The relation between them is :-

T_3 = \dfrac{(T_1 + T_2)}{2}T

3

=

2

(T

1

+T

2

)