Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

i) the maximum height to which it rises

ii) the total time it takes to return to the surface of the Earth. ​

Answer 1

AnswEr:

$\bf{Given}\begin{cases}\sf{Initial\:Velocity,u=49ms^-1}\\\sf{Final\:Velocity,v=0}\\ \sf{g=-9.8ms^-2}\end{cases}$

i) For height we will use the formula

$\bold{\large{\boxed{\sf{\green{h=\frac{v^2-u^2}{2g}}}}}}$

$\longrightarrow \tt \: \frac{0 - {(49)}^{2} }{ - 2 \times 9.8} \\ \\ \longrightarrow \tt \frac{49 \times 49}{19.6} \\ \\ \longrightarrow \tt \: 122.5 \: m$

Therefore, the maximum height to which it rises will be 122.5 m .

____________________________

ii) For the total time it takes to return to the surface of the Earth :

$\longrightarrow \tt \: t = \frac{v - u}{g} \\ \\ \longrightarrow \tt \: \frac{0 - 49}{ - 9.8} \\ \\ \longrightarrow \tt \: 5 \: s \:$

Since, Time of ascend = Time of descend

Total time = Time of ascend + Time of descend

= 2×5

= 10 s ..

Answer 2

$\large{\underline{\underline{\mathrm{\red{Question-}}}}}$

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

i) the maximum height to which it rises

ii) the total time it takes to return to the surface of the Earth.

$\large{\underline{\underline{\mathrm{\red{Answer-}}}}}$

• Maximum height it reaches = 120.05 m.
• The total time it takes to return to the surface of the Earth = 9.8 s

$\large{\underline{\underline{\mathrm{\red{Explanation-}}}}}$

Given :

• Initial velocity ( u ) = 49 m/s
• Final velocity ( v ) = 0

To find :

• The maximum height to which it rises.
• The total time it takes to return to the surface of the Earth.

Solution :

Let's solve it by taking acceleration due to gravity as :

• - 10 m/s²

We know that,

$\huge{\boxed{\rm{\green{v^2\:-\:u^2\:=\:2gS}}}}$

$\implies$ $\rm{S\:=\:\dfrac{v^2\:-\:u^2}{2g}}$

Putting the values,

$\implies$ $\rm{S\:=\:\dfrac{(0)^2-(49)^2}{2\:\times\:(-10)}}$

$\implies$ $\rm{S\:=\:\dfrac{-2401}{-20}}$

$\implies$ $\rm{S\:=\:120.05\:m}$

$\therefore$ Maximum height it reaches = 120.05 m.

$\rule{200}2$

Now, we have to calculate the time taken.

We also know that,

$\huge{\boxed{\rm{\purple{v\:=\:u\:+\:gt}}}}$

$\implies$ $\rm{t\:=\:\dfrac{v\:-\:u}{g}}$

Putting the values,

$\implies$ $\rm{t\:=\:\dfrac{0\:-\:49}{-10}}$

$\implies$ $\rm{t\:=\:\dfrac{-49}{-10}}$

$\implies$ $\rm{t\:=\:4.9\:s}$

But we have to find the total time.

Total time = time for upward direction + time for downward direction

$\implies$ $\rm{Total\:time\:=\:4.9\:+\:4.9}$

$\implies$ $\rm{Total\:time\:=\:9.8\:s}$

$\therefore$ The total time it takes to return to the surface of the Earth = 9.8 s

*Note - You can also solve this question by taking acceleration due to gravity as -9.8 m/s².