Question

A ball is gently dropped from a height of 20m.If its velocity increases uniformly
At the rate of 10 m/s2, with what velocity will it strike the ground? At what time
will it strike the ground?​

Answer 1

Answer:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms ^−2

Time of fall, t=?

We know

v^2  −u^2  =2as

or v ^2  −0=2×10×20=400 or v=20m ^−1

 Now using v=u+at we have

20=0+10×t or t=2s

mrk me brainliest  plzzzz

Answer 2

GIVEN:-

$\large\sf\green{Initial \: velocity \: of\:ball(u)=0}$

$\large\sf\green{Final\:velocity\:of\:ball(v)=?}$

$\large\sf\green{Distance(s)=20m}$

⠀⠀⠀⠀

$\large\sf\pink{Acceleration(a)=10m/s}$

$\large\sf\pink{Time\:of\:Fall(t)=?}$

⠀⠀⠀

$\huge\sf\orange{We\:know,}$

$\longrightarrow$$\large\sf\blue{ {v}^{2} - {u}^{2} = 2as}$

$\longrightarrow$$\large\sf\blue{ {v}^{2} = {(u)}^{2} + 2as}$

$\longrightarrow$$\large\sf\blue{ {v}^{2} = {(0)}^{2} + 2 \times 10 \times 20}$

$\longrightarrow$$\large\sf\blue{{v}^{2}=400}$

$\longrightarrow$$\large\sf\blue{v=√400=20}$

$\longrightarrow$$\large\sf\blue{v=20m/s}$

$\huge\sf\purple{We\:know,}$

$\longrightarrow$$\large\sf{v=u+at}$

$\longrightarrow$$\large\sf{20=0+10×t}$

$\longrightarrow$$\large\sf{10t=20}$

$\longrightarrow$$\large\sf{t=2s}$