Question

A body is thron up with a velocity 50m/s calculate the maximum hight reached and time taken to reach the maximum hight [a=10/s2

Answer 1

Correct Question :-

• A body is thrown up with a velocity 50m/s calculate the maximum height reached and time taken to reach the maximum hight [a=10/s²]

Given :-

• Initial velocity of the body = 50 m /s
• final velocity of the object = 0 m/s

(As the velocity of an object at it's highest point is always zero )

• Acceleration due to gravity = - 10 m /s²

(as the object is moving against gravity )

To find :-

• Maximum height
• Time taken to reach max height

How to solve :-

• As the object accelerates under the influence of gravity ,we can use kinematic equations in order to solve this Question
• We can find maximum height using third equation of motion and time through first equation of motion

Solution :-

As per the third kinematic equation ,

$\bigstar\boxed{\mathtt{\pink{ {v}^{2} = {u}^{2} \underline{+} 2gh}}}$

here ,

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

The value of acceleration of gravity can be positive or negative ,it depends wheather the object is moving along or against gravity (respectively)

• h = maximum height

Now let's substitute the given values in above equation ,

$\implies\mathtt{0 = 2500 - 2 \times 10 \times h}$

$\implies\mathtt{h = \dfrac{2500}{20}}$

$\implies\mathtt{h = 125 \: m}$

The maximum height of the body is 125 m

let's find the value of time by using the first kinematic equation

$\bigstar\boxed{\mathtt{\pink{v = u \underline{ + }gt }}}$

here ,

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

Now let's substitute the given values in above equation ,

$\implies\mathtt{ 0 = 50 - 10 × t }$

$\implies\mathtt{ t \: = \dfrac{50}{10} }$

$\implies\mathtt{ \red{ t \: = 5 \: sec}}$

The time taken by the body to reach maximum height is 5 s

Answer 2

$\;\;\underline{\textbf{\textsf{ Question :-}}}$

•A body is thrown up with a velocity 50m/s calculate the maximum height reached and time taken to reach the maximum hight (a=10/s²)

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

•Initial velocity, u = 50 m /s

•Final velocity, v = 0 m/s

( At highest point)

• Acceleration due to gravity

g = - 10 m /s²

(As the object is moving against gravity)

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Maximum height, H

•Time, t

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{As we know the formula :}}$

$\boxed{\sf v^{2} - u^{2} = 2gH}$

$(\bf 3rd \: kinematic \:equation \: of \: motion)$

$\underline{\:\textsf{Substitute the given values :}}$

$\longrightarrow \mathtt{0 = 2500 - 2 \times 10 \times h}$

$\longrightarrow \mathtt{H = \dfrac{2500}{20}}$

$\longrightarrow \mathtt{H = 125 \: m}$

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Again, we know

$\boxed{\sf v = u + gt}$

$(\bf 1st \: kinematic \:equation \: of \: motion)$

$\underline{\:\textsf{Now, substitute the given values :}}$

$\longrightarrow \mathtt{ 0 = 50 - 10 × t }$

$\longrightarrow \mathtt{ t \: = \dfrac{50}{10} }$

$\longrightarrow \mathtt{ t = 5\: sec}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{The maximum height of the body is \textbf{125 m }}}.$

$\underline{\textsf{The time taken by the body to reach maximum height is \textbf{5 sec}}}.$

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