Question

A ball gently dropped from aheight of 20m.
If its velocity increase uniformly at the
rate of lomls with what velocity will it
strikes the ground. After what time will
it strikes the ground?​

Answer 1

Answer:

HAPPY FRIENDSHIP DAY

Explanation:

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

v2−u2=2as

or v2−0=2×10×20=400 or v=20ms−1

Now using v=u+at we have

20=0+10×t or t=2s

Answer 2

Answer:

as the ball is dropped it's initial velocity is 0

Explanation:

Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’

u =0

s =20 m

a =10 (for sake of easy calculation you can also use 9.8 as both the answer will be the same)

As we know, $V^{2} -u^{2}$ = 2as

$V^{2}$ =  $u^{2}$ + 2as

= 2 x 10 x 20 + 0

= 400

Final velocity of ball, v = 20 m/s

lets use the formula

V = u +at

20  = 0 + 10*t

t = (20-0)/10

= 20/10

= 2 seconds

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