Question

...Please help it's important ❤❤...From the top of a structure 100 m high a body is projected vertically upwards with a velocity of 39.2 m/sec .Find a) how high will it rise b) velocity with which it strikes the ground c) time it takes to reach the ground?​

Answer 1

Answer

Given -

h = 100m

u = 39.2 m/s

a = g = - 9.8 (because body is projected upwards)

where ,

$\longrightarrow$h is height of tower

$\longrightarrow$u is initial velocity.

$\longrightarrow$a is acceleration due to gravity.

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To find -

1. H

2. v

3. t

where

$\longrightarrow$H is the highest height obtained by body.

$\longrightarrow$v is final velocity.

$\longrightarrow$t is time to reach ground.

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Formula used -

Equations of motion -

$\longrightarrow$v = u + at

$\longrightarrow$v² = u² + 2as

$\longrightarrow$s = ut + ½at²

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Solution -

1.

For the highest height obtained by body , final velocity is 0

$\longrightarrow$v = 0

$\longrightarrow$u = 39.2

$\longrightarrow$a = 9.8

By 3rd equation of motion -

v² = u² + 2as

$\longrightarrow$0 = (39.2)² + 2 × (-9.8) H

$\longrightarrow$1536.64 = 19.6H

$\longrightarrow$H = 78.4 m

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2.

Total height = 100 + 78 = 178

$\longrightarrow$h = 178

$\longrightarrow$u = 0

$\longrightarrow$a = g = + 9.8m/s²

By 3rd equation of motion -

v² = u² + 2as

$\longrightarrow$v² = 0 + 2 × 9.8 × 178

$\longrightarrow$v² = 3488.8

$\longrightarrow$v = 59 m/s

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3.

Time taken to reach maximum height -

$\longrightarrow$ u = 39.2 m/s

$\longrightarrow$ a = g = + 9.8m/s²

$\longrightarrow$ v = 0

By 1st equation of motion -

v = u + at

$\longrightarrow$0 = 39.2 - 9.8t

$\longrightarrow$t = 4sec

Time taken to reach ground

$\longrightarrow$ u = 0 m/s

$\longrightarrow$ a = g = + 9.8m/s²

$\longrightarrow$ v = 59

By 1st equation of motion -

v = u + at

59 = 0 + 9.8t

t = 6sec

Total time = 4 + 6 sec

Total time to strike ground = 10sec

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Thanks

Answer 2

☞ Question :

→ From the top of a structure 100 m high a body is projected vertically upwards with a velocity of 39.2 m/s .Find a) how high will it rise b) velocity with which it strikes the ground c) time it takes to reach the ground?

☞ To Find :

• Max. height
• Final velocity
• time taken

☞ Given :

• Height of the building = $100 m$
• Initial velocity = $39.2 ms^{-1}$
• Acceleration due to gravity = $9.8 ms^{-2}$

☞ We know :

➝ Third law of motion :

$\mathtt{\boxed{v^{2} = u^{2} \pm 2gh}}$

→ Where,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• h = height

➝ First law of motion :

$\mathtt{\boxed{v = u \pm gt}}$

→ where,

• v = final Velocity
• u = initial velocity
• g = acceleration due to gravity
• t = time taken

☞ Concept :

➝ From the third law of motion,

$\mathtt{v^{2} = u^{2} \pm 2gh}$

→ we get,

$\mathtt{0 = u^{2} - 2gh}$

$\because$ In this case, the the body is against

the gravity ,hence the g will be Taken as negative.

$\because$ At the highest point , v = 0.

$\mathtt{\Rightarrow -u^{2} = - 2gh}$

$\mathtt{\Rightarrow \cancel{-}u^{2} = \cancel{-}2gh}$

$\mathtt{\Rightarrow u^{2} = 2gh}$

$\mathtt{\Rightarrow \dfrac{u^{2}}{2g} = h}$

→ Thus ,the equation formed is :

$\mathtt{\boxed{\therefore h_{max} = \dfrac{u^{2}}{2g}}}$

➝ The Acceleration due to gravity is taken as positive , when it is falling with the gravity and it is taken as negative when thrown against the gravity .

☞ Solution :

For Finding the Max. Height.

→ We know ,the Formula for finding the Max. height of the body .

$\mathtt{h_{max} = \dfrac{u^{2}}{2g}}$

→ Putting the value in the formula , we get :

$\mathtt{\Rightarrow h_{max} = \dfrac{39.2^{2}}{2 \times 9</p><p>8}}$

$\mathtt{\Rightarrow h_{max} = \dfrac{1536.64}{19.6}}$

$\mathtt{\Rightarrow h_{max} = \cancel{\dfrac{1536.64}{19.6}}}$

$\mathtt{\Rightarrow h_{max} = 78.4 m}$

→ Hence , the Max. height of the body is 78.4 m.

For finding the Final velocity :

• Returning from the highest point , u = 0.

• Total Height = Height of Building + Max. Height

$\Rightarrow 100 m + 78m$

$\Rightarrow 178 m$

→ Using the third law of motion ,

$\mathtt{v^{2} = u^{2} + 2gh}$

$\because$ The body is moving with the gravity , so the g is taken as positive.

→ Putting the value in the formula, we get :

$\mathtt{\Rightarrow v^{2} = 0^{2} + 2 \times 9.8 \times 178}$

$\mathtt{\Rightarrow v^{2} = 3488.8}$

$\mathtt{\Rightarrow v = \sqrt{3488.8}}$

$\mathtt{\Rightarrow v = 59.06 ms^{-1}}$

Hence, the final velocity of the body after reaching the ground is $59.06 ms^{-1}$

For Finding the Total Time of the journey :

• Let the time taken to reach the Max. height From the top of the tower be $t_{1}$.

→ Given,

• v = $0 ms^{-1}$
• u = $39.2 ms^{-1}$
• g = $9.8 ms^{-2}$

Using the first equation of motion,

$\mathtt{v = u - gt}$

$\because$ The body is moving against the gravity , so the g is taken as negative.

→ Putting the value in the equation ,we get,

$\mathtt{\Rightarrow 0 = 39.2 - 9.8t}$

$\mathtt{\Rightarrow - 39.2 = - 9.8t}$

$\mathtt{\Rightarrow \cancel{-} 39.2 = \cancel{-} 9.8t}$

$\mathtt{\Rightarrow 39.2 = 9.8t}$

$\mathtt{\Rightarrow \cancel{39.2} = \cancel{9.8}t}$

$\mathtt{\Rightarrow 4 s = t}$

→ Hence, the time taken to reach the Max. height From the top of the tower is 4 s.

• Let the time taken from the top of the tower to the ground be $t_{2}$

→ Given,

• v = $59.06 ms^{-1}$
• u = $0 ms^{-1}$
• g = $9.8 ms^{-2}$

Using the first equation of motion,

$\mathtt{v = u + gt}$

$\because$ The body is moving with the gravity , so the g is taken alls positive.

→ Putting the value in the equation ,we get,

$\mathtt{\Rightarrow 59.06 = 0 + 9.8t}$

$\mathtt{\Rightarrow 59.06 = 9.8t}$

$\mathtt{\Rightarrow \cancel{59.06} = \cancel{9.8}t}$

$\mathtt{\Rightarrow 6.02 s = t}$

Hence, the time taken from the top of the tower to the ground is 6.02 s.

Total Time:

The time taken to reach the Max. height From the top of the tower + The time taken from the top of the tower to the ground.

$\Rightarrow 4 s + 6.02 s$

$\Rightarrow 10.02 s$

☞ Answer :

Max. height = 78.4 m

Final velocity = 59.06 $ms^{-1}$

Time taken = 10.02 s

☞ Extra Information:

• Second law of motion :

$s = ut + \dfrac{1}{2}at^{2}$

• Law for nth second

$s_{n} = u + \dfrac{1}{2}a(2n - 1)$

Where,

s = displacement

u = initial velocity

v = final velocity

a = acceleration due to gravity

t = time taken