Question

Help me plzz find the answer

Answer 1

Answer:

Let the mass of 1st body be m and the initial velocity be u

So as per Equations of Kinematics :

${v}^{2} = {u}^{2} - 2gh$

$= > {0}^{2} = {u}^{2} - \{ 2 \times g \times (50) \}$

$= > {u}^{2} = 1000$

$= > u = \sqrt{1000} = 31.62 \: m {s}^{ - 1}$

Let mass of second body be 2m and initial velocity be 2u.

${v}^{2} = {(2u)}^{2} - 2g(h2)$

$= > {0}^{2} = 4 {u}^{2} - 2g(h2)$

$= > h2 = \dfrac{4 {u}^{2} }{2g}$

$= > h2 = \dfrac{2 {u}^{2} }{g}$

$= > h2 = \dfrac{2 \times 1000}{10}$

$= > h2 = 200 \: m$

So height reached is 200 m

Answer 2

Answer:

Answer:

Let the mass of 1st body be m and the initial velocity be u

So as per Equations of Kinematics :

{v}^{2} = {u}^{2} - 2ghv

2

=u

2

−2gh

= > {0}^{2} = {u}^{2} - \{ 2 \times g \times (50) \}=>0

2

=u

2

−{2×g×(50)}

= > {u}^{2} = 1000=>u

2

=1000

= > u = \sqrt{1000} = 31.62 \: m {s}^{ - 1}=>u=

1000

=31.62ms

−1

Let mass of second body be 2m and initial velocity be 2u.

{v}^{2} = {(2u)}^{2} - 2g(h2)v

2

=(2u)

2

−2g(h2)

= > {0}^{2} = 4 {u}^{2} - 2g(h2)=>0

2

=4u

2

−2g(h2)

= > h2 = \dfrac{4 {u}^{2} }{2g}=>h2=

2g

4u

2

= > h2 = \dfrac{2 {u}^{2} }{g}=>h2=

g

2u

2

= > h2 = \dfrac{2 \times 1000}{10}=>h2=

10

2×1000

= > h2 = 200 \: m=>h2=200m

So height reached is 200 m