Question

A ball is dropped from the top of the tower. What is its velocity after it has covered 20m? Please answer step by step

Answer 1

Explanation:

A BALL IS DROPPED FROM A HEIGHT,

SO INITIAL VELOCITY (U)=0

ACCELERATION =9.8 M/S^2

DISPLACEMENT =20M

FINAL VELOCITY =V

ACCORDING TO THIRD EQUATION OF MOTION

I.E:

${v}^{2} = {u}^{2} + 2as$

${v}^{2} = {0}^{2} + 2 \times 9.8 \times 20 = > {v}^{2} = 392 = > v = \sqrt{392}$

HERE YOU CAN ALSO TAKE ACCELERATION AS 10M/S^2 IF IN THE QUESTION IT IS NOT SPECIFIED.

Answer 2

Answer:

20m/s

Explanation:

Initial velocity(u): '0' as the object is dropped from rest

Final velocity (v): to be found

distance (S):20m

now S=UT+1/2 AT^2

=> 20=0 X T +1/2 10T^2   [A=accelaration due to gravity 10m/s^2]

=> 20=5t^2

=>t=2 sec

now V=U+AT

=> v=0+10 X 2

=> v=20m/s.........ANS