Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate

of 10 m/s2

, with what velocity will it strike the ground? After what time will it strike the

ground​

Answer 1

u=o,a=10,h=20

h=ut+1/2at^2

t=2& now v=u+1/2at

v=20m/s

Answer 2

Answer:

Figure regards,

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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!?

Provided that:

$\begin{gathered}\begin{gathered}\sf According \: to \: statement \begin{cases} & \sf{Initial \: velocity \: = \bf{0 \: m/s}} \\ \\ & \sf{Final \: velocity \: = \bf{?}} \\ \\ & \sf{Time \: = \bf{?}} \\ \\ & \sf{Height \: = \bf{20 \: metres}} \\ \\ & \sf{Acceleration \: = \bf{10 \: m/s^{2}}}\end{cases}\\ \\\end{gathered}\end{gathered}$

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

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)

Full solution:

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

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

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

$\begin{gathered}:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 20 = 0 + 10(t) \\ \\ :\implies \sf 20 = 0 + 10t \\ \\ :\implies \sf 20 = 10t \\ \\ :\implies \sf 20/10 = \: t \\ \\ :\implies \sf 2 \: = t \\ \\ :\implies \sf t \: = 2 \\ \\ :\implies \sf Time \: = 2 \: seconds\end{gathered}$