Question

A ball is thrown from a height of 20m . Calculate its final velocity before touching the ground.
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Answer 1

Explanation:

Given, s=20m,v=0m/sec,a=9.8m/s

2

1) Initial velocity

v

2

−u

1

=2as

⇒0−u

2

=2(9.8)(20)

u

2

=393m/sec

⇒u=19.8m/sec

2) Final velocity

u=0m/sec

v=?

a=9.8m/s

2

s=20m

v=19.8m/s

3)S

1

=?,t

1

=0.5sec,S

2

=?t

2

=2.5sec

For S

1

=ut

1

+

2

1

a(t

1

)

2

=19.8×0.5+

2

1

(−9.8)(0)

2

=8.675m

For S

2

=ut

2

+

2

1

a(t

2

)

2

=19.8×2.5+

2

1

(−9.8)(2.5)

2

=18.875m

Distance travelled=18.875−8.675=10.2m

4)t=?,u=19.8,s=15m,a=−9.8m/s

2

⇒S=ut+

2

1

(a)(t

2

)

⇒4.9t

2

−19.8t+15=0

⇒t=

2×4.9

19.8±9.8

=1.01s or 3.03s

∴ When going up, t=3.03s

When coming down, t=1.01s