Question

1.(I) A stone is released from the top of a tower of height 19.6 m. Calculate its velocity just before touching the ground. 3 (II) The mass of sun is 2 x 1030 kg and that of earth is 6 x 1024 kg. If the average distance between sun and earth be 1.5x 1011 m, calculate the force of gravitation between them 3 (III) A ball thrown vertically up returns to the thrower after 6s. Find (i) The velocity with which is was thrown up. (ii) The maximum height it reaches. Plz help if answer is right i will make brainlest promise​

Answer 1

$\huge{\underline {\underline{{\gray{ \sf Required \: Answers }}}}}$

$\red{\underline{{ \bf Question }}}$

• A stone is released from the top of a tower of height 19.6 m. Calculate its velocity just before touching the ground

$\red{\underline{{ \bf Answer }}}$

$\underline {\underline{{\purple{ \sf Given }}}}$

• Initial Velocity (u) ➠ 0 m/s
• Height (s) ➠ 19.6 m ➠ 9.8 N
• Acceleration due to gravity (g)

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• Final Velocity

$\underline {\underline{{ \green{\sf Calculating \: Final \: Velocity }}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf{v}^{2} - {u}^{2} = 2gh }}}}$

Substituting Values

☞ v² - (0)² = 2 × 9.8 × 19.6

☞ v² = 384.16

☞ v² = 19.6 m/s

${\gray{ \underbrace{\boxed{\underline{\underline{\orange{ \tt Therefore, \: the \: velocity \: before \: just \: touching \: the \: ground \: is \: 19.6 m }}}}}}}$

$\red{\underline{{ \bf Question }}}$

• The mass of sun is 2 × 10³⁰kg and that of earth is 6 × 10²⁴kg. If the average distance between sun and earth be 1.5 × 10⁻¹¹ m, calculate the force of gravitation between them

$\red{\underline{{ \bf Answer }}}$

$\underline {\underline{{\purple{ \sf Given }}}}$

• Mass of Earth(m₁) ➠ 2 × 10³⁰ kg
• Mass of Sun (m₂) ➠ 6 × 10²⁴ kg
• Distance between sun and earth (r) = 1.5 × 10⁻¹¹ m
• Gravitational Constant (G) =

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• Force of Gravity between them (F)

$\underline {\underline{{ \green{\sf Calculating \: force \: of \: gravity \: between \: them }}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf F = \dfrac{Gm_{1} m_{2}} {r^{2}} }}}}$

Substituting Values

☞ $\sf \dfrac{6.67 \times 10^{-11} \times 2 \times {10}^{ 30} \times 6 \times {10}^{ 24 } } {{(1.5 \times 10^{-11}})^{2} }$

☞ $\sf \dfrac{ 6.67 \times 2 \times 6 \times 10^{(-11+30+24)}}{2.25 \times 10^{22}}$

☞ $\sf \dfrac{ 6.67 \times 2 \times 6 \times (10)^{43}}{2.25 \times 10^{22}}$

☞ $\sf \dfrac{ 6.67 \times 2 \times 6 \times (10)^{43}}{2.25 \times 10^{22}}$

☞ $\sf \dfrac{ 80.04 \times 10^{(43-22)}}{2.25 \times 10^{22}}$

☞ $\sf \dfrac{80.04 \times 10^{21}}{2.25}$

☞ $\sf 35.573 \times 10^{21}$

${\green{ \underbrace{\boxed{\underline{\underline{\pink{ \tt 35.573 \times 10^{21} \: N \: force \: is \: applied }}}}}}}$

$\red{\underline{{ \bf Question }}}$

• A ball thrown vertically up returns to the thrower after 6s. Find (i) The velocity with which is was thrown up. (ii) The maximum height it reaches.

$\red{\underline{{ \bf Answer }}}$

$\underline {\underline{{\purple{ \sf Given }}}}$

• Final Velocity (v) ➠ 0 m/s
• Acceleration due to gravity ➠ 9.8 N

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• The velocity with which is was thrown up i.e Initial Velocity (u)
• The maximum height it reaches i.e Height (h)

Time ball is thrown up and then finally it came down, it means it would take time in going up & coming down

Therefore, 3 second

$\underline {\underline{{ \green{\sf (i) Calculating \: velocity \: with \: which \: it \: was \: thrown \: up}}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf v = u + gt }}}}$

Substituting Values

☞ 0= u -9.8 × 3

☞ - u = -29.4

☞ u = 29.84

${\orange{ \underbrace{\boxed{\underline{\underline{\purple{ \tt \therefore \: It \: was \: thrown \: with \: a velocity \: of \: 29.84 \: m/s }}}}}}}$

$\underline {\underline{{ \green{\sf (ii) Calculating \: the \: maximum \: height \: it \: reaches}}}}$

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf h = ut + \dfrac{1}{2} gt^{2}}}}}$

Substituting Values

☞ h = 29.84 × 3 + ½ × -9.8 × 3²

☞ h = 29.84 × 3 + ½ × -9.8 × 9

☞ h = 89.52 + ½ × -88.2

☞ h = 89.52 - 44.1

☞ h = 45.42

${\purple{ \underbrace{\boxed{\underline{\underline{\red{ \tt Thus, the \: height \: reached \: is \: 45.42 \: m }}}}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\rm\pink{Answer - }$

1.

$\sf\red{Given - }$

$\bf u = 0 m/s$

$\bf a = g = 9.8 m/s^2$

$\bf h = 19.6 m$

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$a is acceleration due to gravity.

$\longrightarrow$h is height of tower.

$\sf\red{To\: find - }$

Final velocity $\longrightarrow$ v

$\rm\pink{Solution - }$

u = 0 m/s

a = g = 9.8 m/s²

h = 19.6 m

By substituting the value in 3rd equation of motion -

$\bf v^2 = u^2 + 2as$

$\implies$$\bf v^2 = 0 + 2 \times 9.8 \times 19.6$

$\implies$$\bf v^2 = 19.6 \times 19.6$

$\implies$$\bf v = 19.6 m/s$

The velocity just before striking the ground is 19.6 m.

━━━━━━━━━━━━━

2.

$\sf\red{Given - }$

$\bf m_s = 2 \times 10^{30}kg$

$\bf m_e = 6 \times 10^{24}kg$

$\bf r = 1.5 \times 10^{11}m$

$\bf G = 6.67 \times 10^{-11}$

where

$\longrightarrow$$\bf m_s$ is the mass of sun.

$\longrightarrow$$\bf m_e$ is the mass of earth.

$\longrightarrow$r is distance between them.

$\longrightarrow$G is universal gravitation constant.

$\sf\red{To \:find - }$

Force of gravitation between them $\longrightarrow$ F

$\sf\red{Formula \:used - }$

$\bf \boxed{ F = \dfrac{Gm_em_s}{r^2} }$

$\rm\pink{Solution -}$

$\bf F = \frac{Gm_em_s}{r^2}$

$\implies$$\bf F = \frac{6.67 \times {10}^{ - 11} \times 2 \times {10}^{30} \times 6 \times {10}^{24} }{( {1.5 \times {10}^{11}) }^{2} }$

$\implies$$\bf F = \frac{ 80 \times {10}^{(30 + 24 - 11)} }{2.25 \times {10}^{22} }$

$\implies$$\bf F = \frac{8 \times {10}^{44} }{2.25 \times {10}^{22} }$(approximately)

$\implies$$\bf F = 3.5 \times {10}^{22}N$

Force of gravitation between sun and earth is $\bf 3.5 \times {10}^{22}N$

━━━━━━━━━━━━━

3.

$\sf\red{Given - }$

a = g = -9.8 m/s (because the object is thrown upwards)

t = 3 sec (the ball is thrown upwards and then came downwards so the time to go upward will be half of total time)

v = 0 m/s ( for max. height )

where

$\longrightarrow$a is acceleration due to gravity.

$\longrightarrow$t is time taken.

$\longrightarrow$v is final velocity.

$\sf\red{To\: find - }$

i) Initial velocity $\longrightarrow$ u

ii) Maximum height $\longrightarrow$ h

$\rm\pink{Solution - }$

i) a = g = -9.8 m/s²

t = 3 sec

v = 0 m/s

Substituting the values in 1st equation of motion -

$\bf v = u + at$

$\implies$$\bf 0 = u - 9.8 \times 3$

$\implies$$\bf u = 29.4 m/s$

The velocity with which the particle was thrown is 29.4 m/s.

━━━━━━━━━━━━━

ii) u = 29.4 m/s

a = -9.8 m/s²

t = 3 sec

where

$\longrightarrow$a is acceleration due to gravity.

$\longrightarrow$t is time taken.

$\longrightarrow$u is initial velocity.

Substituting the value in 2nd equation of motion -

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

$\implies$$\bf s = 29.4 \times 3 - 1/2 \times 9.8 \times 3^2$

$\implies$$\bf s = 88.2 - 44.1$

$\implies$$\bf s = 44.1 m$

Maximum height reached by particle is 44.1 m.

━━━━━━━━━━━━━