Physics, asked by 100anurag1009, 2 months ago

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 10 m s -2, with what velocity will it strike the ground? After what time will it strike the ground?

plss it's urgent

I'll mark the longest answer as brainliest​

Answers

Answered by Anonymous
6

{\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 20 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{20 \: 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)(20) = v^2 - (0)^{2} \\ \\ :\implies \sf 2(10)(20) = v^2 - 0 \\ \\ :\implies \sf 2(10)(20) = v^2 \\ \\ :\implies \sf 2(200) = v^2 \\ \\ :\implies \sf 400 = v^2 \\ \\ :\implies \sf \sqrt{400} = \: v \\ \\ :\implies \sf 20 = \: v \\ \\ :\implies \sf v = \: 20 \\ \\ :\implies \sf Final \: velocity \: = 20 \: m/s

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

:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 20 = 0 + 10(t) \\ \\ :\implies \sf 20 = 0 + 10t \\ \\ :\implies \sf 20 = 10t \\ \\ :\implies \sf 20/10 = \: t \\ \\ :\implies \sf 2 \: = t \\ \\ :\implies \sf t \: = 2 \\ \\ :\implies \sf Time \: = 2 \: seconds

Answered by anupamsgpgi
0

Answer:

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