Question

Answer the following question

Answer 1

Given :-

• Time taken = 5 seconds
• g= 10m/s^2

To find :-

• Maximum height and Initial speed

SOLUTION :-

First we find initial velocity of the object

As we know that ,

v = u +gt

Here v=0 and g = -10

0 = u + (-10)(5)

u-50 = 0

u = 50 m/s

So the initial velocity of the body is 50m/s

Now, we have formula to find the maximum height that is

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

Substituting the values

${h_{max}}= \dfrac{(50)^2}{2(10)}$

$=\dfrac{2500}{20}$

$=\dfrac{250}{2}$

$= 125m$

So, maximum height is 125m and initial speed is 50m/s

Reasons to take the value of g is -10 and v is 0

As the ball is thrown upwards and reaches highest point means It stops So, the value of "v" will be 0

As we know that direction of "g" is always downwords that is + 10 but It thrown upwards that means direction of "g" is changes i.e (upwards) So, the value of g is -10

________________

Know more some more formulae :-

• v² - u² = 2as
• s = ut + 1/2 at²
• s = vt - 1/2 at²
• v²-u²/2s = a
• a = v-u/t

t = time taken

s = Distance covered

v= final velocity

a = accelartion

Answer 2

${\large{\pmb{\sf{\underline{RequirEd \: Solution...}}}}}$

${\bigstar \:{\pmb{\sf{\underline{About \: question...}}}}}$

⋆ Question: A ball is thrown upwards with a speed such that it reaches the highest point in 5 seconds. What is the initial speed and maximum height reached? Take g = 10 m/s²

⋆ Understanding the question: This question says that we have to find out the initial speed and maximum height reached by the ball if it is thrown upwards with a speed such that it reaches the highest point in 5 seconds. It is also given that we have to take g as 10 metre per second sq. Let's solve this question!

${\bigstar \:{\pmb{\sf{\underline{Provided \: that...}}}}}$

$\sf According \: to \: statement \begin{cases} & \sf{Final \: velocity \: = \bf{0 \: ms^{-1}}} \\ \\ & \sf{Initial \: velocity \: = \bf{?}} \\ \\ & \sf{Maximum \: height \: = \bf{?}} \\ \\ & \sf{Time \: taken \: = \bf{5 \: seconds}} \\ \\ & \sf{g \: = \bf{-10 \: ms^{-2}}} \end{cases}$

Don't be confused!

⋆ We take the value of g as -10 m/s² expect of +10 m/s² because the ball is thrown in upward direction that's why the sign changed!

⋆ Final velocity came as zero because when the ball is thrown then it stopped at a highest point!

${\bigstar \:{\pmb{\sf{\underline{Using \: dimensions...}}}}}$

⋆ First equation of motion is given by

• ${\small{\underline{\boxed{\sf{v \: = u \: + gt}}}}}$

⋆ Third equation of motion is given by

• ${\small{\underline{\boxed{\sf{2gs \: = v^2 \: - u^2}}}}}$

(Where, v denotes final velocity , u denotes initial velocity , t denotes time taken , g denotes acceleration due to gravity , s denotes displacement or distance or height)

${\bigstar \:{\pmb{\sf{\underline{Full \: Solution...}}}}}$

~ Firstly by using first equation of motion let us find out the initial velocity!

$:\implies \sf v \: = u \: + gt \\ \\ :\implies \sf 0 = u + (-10)(5) \\ \\ :\implies \sf 0 = u + (-50) \\ \\ :\implies \sf 0 = u - 50 \\ \\ :\implies \sf 0 + 50 = \: u \\ \\ :\implies \sf 50 \: = u \\ \\ :\implies \sf u \: = 50 \: ms^{-1} \\ \\ :\implies \sf Initial \: velocity \: = 50 \: ms^{-1}$

~ Now by using third equation of motion let us find out the maximum height!

$:\implies \sf 2gs \: = v^2 \: - u^2 \\ \\ :\implies \sf 2(-10)(s) = 0^2 - 50^{2} \\ \\ :\implies \sf 2(-10)(s) = 0 - 2500 \\ \\ :\implies \sf 2(-10)(s) = -2500 \\ \\ :\implies \sf 2(-10s) = -2500 \\ \\ :\implies \sf -20s = -2500 \\ \\ :\implies \sf 20s = 2500 \\ \\ :\implies \sf s \: = 2500/20 \\ \\ :\implies \sf s \: = 250/2 \\ \\ :\implies \sf s \: = 125 \: metres \\ \\ :\implies \sf Height \: = 125 \: metres$

∴ Initial velocity is 50 metre per second & the height is 125 metres.