Question

A ball is gently dropped from a height of 125 m if it's Velocity increase uniformly at the rate of 10 m/ sec^ 2 with velocity will it is strike the ground? after what time it strike the ground​

Answer 1

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

${\bigstar \:{\pmb{\sf{\underline{Understanding \: the \: question...}}}}}$

⋆ This question says that there is a ball that is gently dropped from a height that is of 125 metres, now this question says that if it's velocity increase uniformly at the rate of 10 metres per second sq. then with velocity will it is strike the ground and it is also asked that after what time the ball will strike or hit the ground!?

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

$\sf According \: to \: statement \begin{cases} & \sf{Initial \: velocity \: = \bf{0 \: m/s}} \\ \\ & \sf{Final \: velocity \: = \bf{?}} \\ \\ & \sf{Time \: = \bf{?}} \\ \\ & \sf{Height \: = \bf{125 \: metres}} \\ \\ & \sf{Acceleration \: = \bf{10 \: m/s^{2}}}\end{cases}$

Don't be confused! Initial velocity comes as zero because the ball is gently dropped from a certain height!

${\bigstar \:{\pmb{\sf{\underline{Knowledge \: required...}}}}}$

⋆ First equation of motion =

⠀⠀⠀${\small{\underline{\boxed{\sf{\rightarrow v \: = u \: + at}}}}}$

⋆ Third equation of motion =

⠀⠀⠀${\small{\underline{\boxed{\sf{\rightarrow 2as \: = v^2 - \: u^2}}}}}$

(Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration , t denotes time , s denotes displacement or distance or height)

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

~ Firstly by using third equation of motion let us find out the final velocity.

$:\implies \sf 2as \: = v^2 - \: u^2 \\ \\ :\implies \sf 2(10)(125) = v^2 - 0^2 \\ \\ :\implies \sf 2(1250) = v^2 - 0 \\ \\ :\implies \sf 2(1250) = v^2 \\ \\ :\implies \sf 2500 = v^2 \\ \\ :\implies \sf \sqrt{2500} = v \\ \\ :\implies \sf 50 = v \\ \\ :\implies \sf v = 50 \\ \\ :\implies \sf Final \: velocity \: = 50 \: m/s$

Henceforth, final velocity is 50 m/s

~ Now let's find out the time that the ball will strike the ground by using first equation of motion.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 50 = 0 + 10(t) \\ \\ :\implies \sf 50 = 10(t) \\ \\ :\implies \sf 50 = 10t \\ \\ :\implies \sf 50/10 = t \\ \\ :\implies \sf 5 = t \\ \\ :\implies \sf t = 5 \\ \\ :\implies \sf Time \: = 5 \: seconds$

Henceforth, 5 seconds is the time when the ball will hit the ground.